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Add and subtract rational numbers

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5121667
Evaluate. a) \(-14-19\) b) \(22-(-11)\) c) \(-35-(-15)\) d) \(8-23\)

Hints

- Rewrite subtraction as adding the opposite. - Decide the sign before calculating. - Picture the movement on a number line.

Solution

1. \(-14-19=-33\). 2. Subtracting a negative is adding its opposite: \(22-(-11)=22+11=33\). 3. \(-35-(-15)=-35+15=-20\). 4. \(8-23=-15\).

Answer

a) \(-33\) b) \(33\) c) \(-20\) d) \(-15\)
5121727
A submarine is at \(-150\,\text{m}\) relative to sea level and then descends another \(75\,\text{m}\). Write an integer addition equation and find its new position.

Hints

- Positions below sea level are negative. - Descending makes the position more negative. - Represent the change as a signed number.

Solution

1. The starting position is \(-150\,\text{m}\). 2. Descending \(75\,\text{m}\) is represented by adding \(-75\). 3. \(-150+(-75)=-225\), so the new position is \(-225\,\text{m}\).

Answer

\(-150+(-75)=-225\). The submarine is at \(-225\,\text{m}\).
5128327
Evaluate each expression and simplify the result. a) \(45-72\) b) \(-18+(-12)\) c) \(\frac{5}{6}-\frac{1}{6}\) d) \(-\frac{3}{8}-\frac{1}{8}\)

Hints

- Determine the sign before calculating. - When adding two negative numbers, add their absolute values and keep the negative sign. - With like denominators, combine the numerators. - Simplify fraction results.

Solution

1. For a), \(45-72=-27\). 2. For b), \(-18+(-12)=-30\). 3. For c), \(\frac{5}{6}-\frac{1}{6}=\frac{4}{6}=\frac{2}{3}\). 4. For d), \(-\frac{3}{8}-\frac{1}{8}=-\frac{4}{8}=-\frac{1}{2}\).

Answer

a) \(-27\) b) \(-30\) c) \(\frac{2}{3}\) d) \(-\frac{1}{2}\)
5180887
A weather station records a temperature of \(-9\,^\circ\text{F}\) one winter morning. During the morning, the temperature rises by \(14\,^\circ\text{F}\). What temperature does the station now show?

Hints

- Picture a thermometer as a vertical number line. - Which direction represents an increase in temperature? - First determine how far it is from \(-9\) to \(0\).

Solution

1. Represent the increase by adding a positive number: \(-9+14\). 2. Calculate: \(-9+14=5\), so the temperature is \(5\,^\circ\text{F}\).

Answer

The station now shows \(5\,^\circ\text{F}\).
5180897
Mr. Weber's checking account has a balance of \(-\$180\). He deposits \(\$250\) in cash. What is the new account balance?

Hints

- A negative account balance represents money owed to the bank. - What happens to the amount owed when money is deposited? - Will the account still have a negative balance after the deposit?

Solution

1. Add the deposit to the current balance: \(-180+250\). 2. Calculate: \(-180+250=70\).

Answer

The new account balance is \(\$70\).
5181187
A type of plastic loses its flexibility and becomes brittle at \(-35\,^\circ\text{C}\). A new material mixture raises this threshold by \(12\) degrees. At what temperature does the new plastic become brittle?

Hints

- Picture the temperatures on a thermometer or number line. - Does raising a temperature threshold make the value greater or less? - Which direction do you move on the number line when a temperature increases?

Solution

1. The original threshold is \(-35\,^\circ\text{C}\). 2. Raising the threshold by \(12\) degrees means adding \(12\): \(-35+12=-23\).

Answer

The new plastic becomes brittle at \(-23\,^\circ\text{C}\).
5181197
A hiker begins a climb in a valley where the temperature is \(4\,^\circ\text{F}\). By the time the hiker reaches a mountain shelter, the temperature has dropped a total of \(15\) degrees. What is the temperature at the shelter?

Hints

- Decide whether the final temperature should be positive or negative. - You are subtracting a number greater than the starting value. - A thermometer sketch can help you cross zero.

Solution

1. A drop of \(15\) degrees means subtracting \(15\) from the initial temperature. 2. Calculate: \(4-15=-11\), so the temperature is \(-11\,^\circ\text{F}\).

Answer

The temperature at the shelter is \(-11\,^\circ\text{F}\).
5182807
What number is \(58\) greater than \(-33\)? Write and evaluate an equation.

Hints

- Increasing a number means moving right on a number line. - Translate “greater than” into addition. - Check whether the result should be positive or negative.

Solution

1. “\(58\) greater than” means add \(58\) to \(-33\). 2. \(-33+58=25\).

Answer

\(-33+58=25\)
5100477
Evaluate the expression. Show the value of the quantity in parentheses before finding the final value. \(7 \frac{5}{12} - \left(3 \frac{5}{6} + 2 \frac{1}{2}\right)\)

Hints

- Evaluate the grouped expression before doing the outside subtraction. - Use a common denominator when adding the fractional parts. - Keep the exact fractional value of the grouped expression for the final subtraction.

Solution

1. Evaluate the parentheses: \(3\frac{5}{6}+2\frac{1}{2}=3\frac{5}{6}+2\frac{3}{6}=6\frac{1}{3}\). 2. Rewrite \(6\frac{1}{3}=6\frac{4}{12}\). 3. Subtract: \(7\frac{5}{12}-6\frac{4}{12}=1\frac{1}{12}\).

Answer

Quantity in parentheses: \(6\frac{1}{3}\) Final value: \(1\frac{1}{12}\)
5100487
Which number is exactly halfway between \(-2.4\) and \(3.6\)? a) \(0.6\) b) \(0.8\) c) \(1.0\) d) \(1.2\)

Hints

- The midpoint is the average of the two endpoints. - Add the two values first. - Divide the sum by \(2\).

Solution

1. Find the midpoint by adding the two numbers and dividing by \(2\): \(\frac{-2.4+3.6}{2}\). 2. The sum is \(-2.4+3.6=1.2\). 3. Divide by \(2\): \(1.2\div2=0.6\).

Answer

a) \(0.6\)
5103587
Evaluate each expression. Pay attention to signs and grouping symbols. a) \(-18+42-(-15)\) b) \(125-(250-75)\) c) \(-15-[-40+(12-22)]\)

Hints

- Subtracting a negative is equivalent to adding its opposite. - Work from the innermost grouping symbols outward. - Simplify signs before doing the arithmetic when helpful.

Solution

1. For a), \(-18+42-(-15)=-18+42+15=39\). 2. For b), evaluate the parentheses first: \(250-75=175\). Then \(125-175=-50\). 3. For c), \(12-22=-10\), then \(-40+(-10)=-50\), and finally \(-15-(-50)=35\).

Answer

a) \(39\) b) \(-50\) c) \(35\)
5106197
Calculate the value of the expression and write the result in simplest form: \(\frac{5}{12} - \left(\frac{1}{4} + \frac{1}{3}\right)\)

Hints

- Which part of the expression should you evaluate first? - How can you rewrite fractions with different denominators so they have a common denominator? - Check whether the final fraction can be simplified.

Solution

1. Evaluate the parentheses first: \(\frac{1}{4} + \frac{1}{3} = \frac{3}{12} + \frac{4}{12} = \frac{7}{12}\). 2. Subtract: \(\frac{5}{12} - \frac{7}{12} = -\frac{2}{12}\). 3. Simplify: \(-\frac{2}{12} = -\frac{1}{6}\).

Answer

\(-\frac{1}{6}\)
5106207
Calculate: \(\left(\frac{3}{8} - \frac{5}{6}\right) + \frac{11}{12}\)

Hints

- First rewrite the fractions inside the parentheses with a common denominator. - What sign should the result have when the larger fraction is subtracted from the smaller fraction? - How do you add a negative fraction and a positive fraction?

Solution

1. Evaluate the parentheses using a common denominator of \(24\): \(\frac{3}{8} - \frac{5}{6} = \frac{9}{24} - \frac{20}{24} = -\frac{11}{24}\). 2. Rewrite \(\frac{11}{12}\) as \(\frac{22}{24}\). 3. Add: \(-\frac{11}{24} + \frac{22}{24} = \frac{11}{24}\).

Answer

\(\frac{11}{24}\)
5106287
Calculate each expression. Write each answer as a fraction in simplest form or as a mixed number. a) \(3 \frac{5}{12}+1 \frac{1}{4}\) b) \(6 \frac{2}{5}-4 \frac{7}{10}\) c) \(2 \frac{3}{4}+0.125\)

Hints

- Rewrite fractional parts with common denominators before adding or subtracting. - For the decimal, consider an equivalent fraction. - If the fractional part of a mixed number is too small to subtract, regroup one whole.

Solution

1. For a), rewrite \(1 \frac{1}{4}\) as \(1 \frac{3}{12}\). Then \(3 \frac{5}{12}+1 \frac{3}{12}=4 \frac{8}{12}=4 \frac{2}{3}\). 2. For b), rewrite \(6 \frac{2}{5}\) as \(6 \frac{4}{10}\). Regroup one whole: \(6 \frac{4}{10}=5 \frac{14}{10}\). Then \(5 \frac{14}{10}-4 \frac{7}{10}=1 \frac{7}{10}\). 3. For c), rewrite \(0.125\) as \(\frac{1}{8}\) and \(2 \frac{3}{4}\) as \(2 \frac{6}{8}\). Then \(2 \frac{6}{8}+\frac{1}{8}=2 \frac{7}{8}\).

Answer

a) \(4 \frac{2}{3}\) b) \(1 \frac{7}{10}\) c) \(2 \frac{7}{8}\)
5106377
Calculate each expression. Write each answer as a mixed number in simplest form. a) \(3 \frac{2}{5}+4 \frac{5}{6}\) b) \(8 \frac{1}{4}-5 \frac{5}{8}\) c) \(10-\left(2 \frac{1}{3}+4 \frac{3}{5}\right)\)

Hints

- Rewrite fractions with unlike denominators using a common denominator. - In a mixed-number subtraction, you may need to regroup one whole. - Evaluate expressions inside parentheses before the rest of the calculation. - A whole number can be rewritten as a fraction or mixed number when that makes the subtraction easier.

Solution

1. For a), use denominator \(30\): \(3 \frac{12}{30}+4 \frac{25}{30}=7 \frac{37}{30}=8 \frac{7}{30}\). 2. For b), rewrite \(8 \frac{1}{4}\) as \(8 \frac{2}{8}\), then regroup: \(7 \frac{10}{8}-5 \frac{5}{8}=2 \frac{5}{8}\). 3. For c), evaluate the parentheses: \(2 \frac{1}{3}+4 \frac{3}{5}=2 \frac{5}{15}+4 \frac{9}{15}=6 \frac{14}{15}\). Then \(10-6 \frac{14}{15}=3 \frac{1}{15}\).

Answer

a) \(8 \frac{7}{30}\) b) \(2 \frac{5}{8}\) c) \(3 \frac{1}{15}\)
5106527
Rearrange and group the terms to make the calculation efficient. Write the equivalent regrouped expression you use, then evaluate it. \(5\frac{3}{8} + 4\frac{2}{5} - 2\frac{3}{8} + 1\frac{3}{5}\)

Hints

- Keep each sign attached to its term when changing the order. - Look for terms whose fractional parts combine without introducing a new denominator. - Your response should show the regrouping, not only the final number.

Solution

1. Rewrite subtraction as addition of the opposite and regroup: \(\left(5\frac{3}{8}-2\frac{3}{8}\right)+\left(4\frac{2}{5}+1\frac{3}{5}\right)\). 2. The groups are \(3\) and \(6\). 3. Therefore, the value is \(9\).

Answer

One efficient regrouping is \(\left(5\frac{3}{8}-2\frac{3}{8}\right)+\left(4\frac{2}{5}+1\frac{3}{5}\right)\), which equals \(9\).
5106617
For each expression, rewrite any fraction as a terminating decimal before evaluating. If an expression has no fraction, state that no conversion is needed. a) \(0.25 + \frac{1}{2}\) b) \(0.4 - \frac{3}{5}\) c) \(0.75 - 1\) d) \(1.1 + \frac{9}{10}\)

Hints

- Write each fraction with a denominator of \(10\) or \(100\) when possible. - Keep the converted decimal visible in your response before evaluating. - Check the sign of each result against the relative sizes of the terms.

Solution

1. a) \(\frac{1}{2}=0.5\), so \(0.25+0.5=0.75\). 2. b) \(\frac{3}{5}=0.6\), so \(0.4-0.6=-0.2\). 3. c) No conversion is needed: \(0.75-1=-0.25\). 4. d) \(\frac{9}{10}=0.9\), so \(1.1+0.9=2\).

Answer

a) \(\frac{1}{2}=0.5\); result \(0.75\) b) \(\frac{3}{5}=0.6\); result \(-0.2\) c) No fraction is present, so no conversion is needed; result \(-0.25\) d) \(\frac{9}{10}=0.9\); result \(2\)
5106827
Lucas and Mia evaluate \(0.125 + \frac{3}{4} + 0.25 + \frac{1}{8}\) in different ways. Lucas converts the fractions to decimals. Mia converts the decimals to fractions and rearranges the addends. Carry out both methods. Which method seems more efficient here? Explain.

Hints

- Convert only values whose equivalents you know exactly. - Look for pairs that combine to a whole number. - Rearranging addends does not change their sum.

Solution

1. Lucas’s method: \(\frac{3}{4} = 0.75\) and \(\frac{1}{8} = 0.125\). Then \(0.125 + 0.75 + 0.25 + 0.125 = 1.25\). 2. Mia’s method: \(0.125 = \frac{1}{8}\) and \(0.25 = \frac{1}{4}\). Then \(\frac{1}{8} + \frac{3}{4} + \frac{1}{4} + \frac{1}{8}\). 3. Regroup: \(\left(\frac{1}{8} + \frac{1}{8}\right) + \left(\frac{3}{4} + \frac{1}{4}\right) = \frac{1}{4} + 1 = 1.25\). 4. Both methods are efficient. Mia’s grouping immediately creates a whole number from the fourths.

Answer

Both methods give \(1.25\). Mia’s method is especially efficient because \(\frac{3}{4} + \frac{1}{4} = 1\).
5106837
Evaluate \(\frac{5}{6} + 0.4 + \frac{1}{6} + 1.1\). Decide whether to convert every value to fractions or decimals, or to combine compatible terms separately. Briefly justify your choice.

Hints

- Compare the work created by converting the sixths to decimals with the work created by keeping them as fractions. - Look for terms that can combine exactly without introducing a new representation. - Your justification should refer to the actual numbers in this expression.

Solution

1. Converting the sixths to decimals would create repeating decimals, so keep them as fractions. 2. Group the fractions: \(\frac{5}{6} + \frac{1}{6} = 1\). 3. Group the decimals: \(0.4 + 1.1 = 1.5\). 4. Add the partial sums: \(1 + 1.5 = 2.5\).

Answer

One particularly efficient choice is to combine the fractions and decimals separately. Converting the sixths to decimals would introduce repeating decimals, while \(\frac{5}{6}+\frac{1}{6}=1\) exactly and \(0.4+1.1=1.5\), so the value is \(2.5\). Another exact strategy is acceptable if it is shown correctly and the efficiency justification refers to the actual conversions and arithmetic required.
5106887
Evaluate the expression by converting the fractions to terminating decimals and then showing one regrouping that creates convenient decimal pairs. \(\frac{2}{5} + 1.35 - \frac{3}{4} + 0.1\)

Hints

- Convert each fraction exactly before doing the addition or subtraction. - Keep each sign attached to its term when regrouping. - Show the regrouped decimal expression in your answer.

Solution

1. Convert: \(\frac{2}{5}=0.4\) and \(\frac{3}{4}=0.75\). 2. Regroup: \((0.4+0.1)+(1.35-0.75)\). 3. Evaluate: \(0.5+0.6=1.1\).

Answer

\(\frac{2}{5}=0.4\), \(\frac{3}{4}=0.75\), and \((0.4+0.1)+(1.35-0.75)=1.1\).
5112277
Evaluate the expression and write the result as a decimal: \(\frac{7}{10}-\left(\frac{2}{5}+\frac{1}{2}\right)\)

Hints

- Evaluate the parentheses first. - Fractions need a common denominator before you add them. - How do you write tenths as decimals?

Solution

1. Add inside the parentheses: \(\frac{2}{5}+\frac{1}{2}=\frac{4}{10}+\frac{5}{10}=\frac{9}{10}\). 2. Subtract: \(\frac{7}{10}-\frac{9}{10}=-\frac{2}{10}\). 3. Convert to a decimal: \(-\frac{2}{10}=-0.2\).

Answer

\(-0.2\)
5112757
Find the error in the calculation and give the correct result. \(7\frac{1}{4} - 2.5 = (7 - 2) + (0.25 + 0.5) = 5 + 0.75 = 5.75\)

Hints

- The entire second number must be subtracted. - Track the subtraction sign when decomposing a number into whole-number and decimal parts. - Rewrite both numbers as decimals or both as fractions.

Solution

1. The error is that \(0.5\), which is part of the number being subtracted, was added instead of subtracted. 2. Convert the mixed number: \(7\frac{1}{4} = 7.25\). 3. Subtract correctly: \(7.25 - 2.5 = 4.75\).

Answer

The \(0.5\) was added instead of subtracted. The correct calculation is \(7.25 - 2.5 = 4.75\).
5112817
For each expression, choose a fraction or decimal form that avoids unnecessary work. Write the conversion or rewrite you chose before evaluating. a) \(\frac{3}{4} + 0.15\) b) \(0.8 - \frac{1}{5}\) c) \(-\frac{2}{3} + 0.5\) d) \(1.125 - \frac{1}{8}\)

Hints

- Ask whether the fraction has a short terminating decimal before converting it. - A repeating decimal is often less convenient than keeping a fraction exact. - Show the representation change you actually use.

Solution

1. a) \(\frac{3}{4}=0.75\), so \(0.75+0.15=0.9\). 2. b) \(\frac{1}{5}=0.2\), so \(0.8-0.2=0.6\). 3. c) \(0.5=\frac{1}{2}\), so \(-\frac{2}{3}+\frac{1}{2}=-\frac{1}{6}\). 4. d) \(\frac{1}{8}=0.125\), so \(1.125-0.125=1\).

Answer

a) \(\frac{3}{4}=0.75\); result \(0.9\) b) \(\frac{1}{5}=0.2\); result \(0.6\) c) \(0.5=\frac{1}{2}\); result \(-\frac{1}{6}\) d) \(\frac{1}{8}=0.125\); result \(1\)
5112877
Evaluate each expression. For every part, first write the exact fraction-to-decimal conversion you use; then calculate. a) \(0.45 + \frac{3}{20}\) b) \(\frac{5}{8} - 0.125\) c) \(-1.2 + \frac{1}{5}\)

Hints

- Rewrite each fraction with denominator \(10\), \(100\), or \(1000\) when possible. - Keep the conversion visible before combining the values. - Check the sign of the final sum or difference.

Solution

1. a) \(\frac{3}{20}=0.15\), so \(0.45+0.15=0.6\). 2. b) \(\frac{5}{8}=0.625\), so \(0.625-0.125=0.5\). 3. c) \(\frac{1}{5}=0.2\), so \(-1.2+0.2=-1\).

Answer

a) \(\frac{3}{20}=0.15\); \(0.45+0.15=0.6\) b) \(\frac{5}{8}=0.625\); \(0.625-0.125=0.5\) c) \(\frac{1}{5}=0.2\); \(-1.2+0.2=-1\)
5113207
Before evaluating, show an equivalent rearrangement and regrouping that creates two whole-number partial sums. Then evaluate and name the properties that justify changing the order and grouping. \(3.7 + \frac{2}{3} + 1.3 + 1\frac{1}{3}\)

Hints

- Keep each addend unchanged when you reorder the sum. - Look for pairs that produce especially simple partial sums. - Name the properties that permit changing order and grouping.

Solution

1. Use the commutative and associative properties of addition to write \((3.7+1.3)+\left(\frac{2}{3}+1\frac{1}{3}\right)\). 2. The decimal group is \(5\). 3. The fraction group is \(\frac{2}{3}+\frac{4}{3}=2\). 4. Therefore, the sum is \(7\).

Answer

A valid regrouping is \((3.7+1.3)+\left(\frac{2}{3}+1\frac{1}{3}\right)=5+2=7\). The commutative and associative properties of addition justify the changed order and grouping.
5117107
Evaluate each expression. a) \(-4.8-2.35\) b) \(-\frac{5}{6}+\frac{1}{3}\) c) \(0.75+(-1.2)\) d) \(-\left(\frac{1}{4}-\frac{5}{8}\right)\)

Hints

- Pay attention to signs when removing parentheses. - Use common denominators before adding or subtracting fractions. - Before calculating with decimals, decide whether the result should be positive or negative.

Solution

1. For a), \(-4.8-2.35=-7.15\). 2. For b), \(-\frac{5}{6}+\frac{1}{3}=-\frac{5}{6}+\frac{2}{6}=-\frac{1}{2}\). 3. For c), \(0.75+(-1.2)=-0.45\). 4. For d), \(\frac{1}{4}-\frac{5}{8}=\frac{2}{8}-\frac{5}{8}=-\frac{3}{8}\). The leading negative changes the sign, giving \(\frac{3}{8}\).

Answer

a) \(-7.15\) b) \(-\frac{1}{2}\) c) \(-0.45\) d) \(\frac{3}{8}\)
5117167
Evaluate the expression: \((-4.2+1.7)-(0.8-2.3)\)

Hints

- Evaluate both sets of parentheses first. - What happens when you subtract a negative number? - A number line can help you check the final sign.

Solution

1. First parentheses: \(-4.2+1.7=-2.5\). 2. Second parentheses: \(0.8-2.3=-1.5\). 3. Subtract: \(-2.5-(-1.5)=-1\).

Answer

\(-1\)
5117947
Calculate the value of the expression and write the result as a mixed number: \(\left(5 \frac{3}{4}-1 \frac{1}{2}\right)+2 \frac{5}{8}\)

Hints

- Evaluate the expression inside the parentheses first. - Rewrite the fractional parts with common denominators before subtracting or adding. - Simplify the result inside the parentheses before continuing. - Convert the intermediate result to eighths before the final addition.

Solution

1. Evaluate the parentheses: \(5 \frac{3}{4}-1 \frac{1}{2}=5 \frac{3}{4}-1 \frac{2}{4}=4 \frac{1}{4}\). 2. Rewrite \(4 \frac{1}{4}\) as \(4 \frac{2}{8}\). 3. Add: \(4 \frac{2}{8}+2 \frac{5}{8}=6 \frac{7}{8}\).

Answer

\(6 \frac{7}{8}\)
5117987
A point starts at \(-2.3\) on a number line. It moves \(4.5\) units to the right and then \(1.4\) units to the left. Where does the point end?

Hints

- A move to the right corresponds to addition. - A move to the left corresponds to subtraction. - Carry out the moves in order.

Solution

1. Moving right means add: \(-2.3+4.5=2.2\). 2. Moving left means subtract: \(2.2-1.4=0.8\). 3. The point ends at \(0.8\).

Answer

\(0.8\)
5118707
Evaluate and write the result as a fraction in simplest form: \(\frac{2}{5}+\frac{1}{2}-\frac{3}{10}\)

Hints

- Find a common denominator for \(5\), \(2\), and \(10\). - Rewrite all three fractions before adding and subtracting. - Simplify the final fraction.

Solution

1. Use denominator \(10\): \(\frac{2}{5}=\frac{4}{10}\) and \(\frac{1}{2}=\frac{5}{10}\). 2. Calculate: \(\frac{4}{10}+\frac{5}{10}-\frac{3}{10}=\frac{6}{10}=\frac{3}{5}\).

Answer

\(\frac{3}{5}\)
5121607
Evaluate each sum or difference of rational numbers. a) \(-8.7+(-13.5)\) b) \(4.2-9.8\) c) \(-\frac{5}{9}+\frac{1}{6}\) d) \(\frac{2}{3}-\frac{3}{4}+\left(-\frac{1}{2}\right)\)

Hints

- What sign results when two negative numbers are added? - For unlike signs, compare absolute values. - Use a common denominator for fractions. - You can rewrite subtraction as addition of the opposite.

Solution

1. For a), \(-8.7+(-13.5)=-22.2\). 2. For b), \(4.2-9.8=-5.6\). 3. For c), use denominator \(18\): \(-\frac{10}{18}+\frac{3}{18}=-\frac{7}{18}\). 4. For d), use denominator \(12\): \(\frac{8}{12}-\frac{9}{12}-\frac{6}{12}=-\frac{7}{12}\).

Answer

a) \(-22.2\) b) \(-5.6\) c) \(-\frac{7}{18}\) d) \(-\frac{7}{12}\)
5121677
Evaluate each expression. Write the result as either a decimal or a fraction. a) \(\frac{3}{4}-1.25\) b) \(-0.6-\left(-\frac{1}{5}\right)\) c) \(2-\frac{1}{2}-0.75\) d) \(-\frac{1}{8}-0.125\)

Hints

- Choose fractions or decimals and convert values to the same form. - Recall common fraction-decimal equivalents such as \(\frac{1}{2}\) and \(\frac{1}{5}\). - Pay close attention to subtracting a negative number.

Solution

1. For a), \(0.75-1.25=-0.5=-\frac{1}{2}\). 2. For b), \(-0.6-(-0.2)=-0.4=-\frac{2}{5}\). 3. For c), \(2-0.5-0.75=0.75=\frac{3}{4}\). 4. For d), \(-0.125-0.125=-0.25=-\frac{1}{4}\).

Answer

a) \(-0.5\), or \(-\frac{1}{2}\) b) \(-0.4\), or \(-\frac{2}{5}\) c) \(0.75\), or \(\frac{3}{4}\) d) \(-0.25\), or \(-\frac{1}{4}\)
5121687
The expressions are \(A=x-5.4\) and \(B=-3.2-y\). Evaluate them for \(x=2.1\) and \(y=-1.8\). Which value is greater?

Hints

- Substitute each given value into its expression. - Subtracting a negative number is equivalent to adding its opposite. - On a number line, the number farther to the right is greater.

Solution

1. \(A=2.1-5.4=-3.3\). 2. \(B=-3.2-(-1.8)=-3.2+1.8=-1.4\). 3. Since \(-1.4>-3.3\), \(B\) has the greater value.

Answer

\(A=-3.3\) and \(B=-1.4\). The value of \(B\) is greater.
5121697
During a winter night, the temperature drops to \(-12\,^\circ\text{C}\). By noon the next day, the temperature has risen to \(7\,^\circ\text{C}\). By how many degrees Celsius did the temperature increase?

Hints

- Picture the temperatures on a number line. - Find the distance from the negative temperature to zero. - Find the distance from zero to the positive temperature. - Which operation finds the change from an initial value to a final value?

Solution

1. The initial temperature is \(-12\,^\circ\text{C}\). 2. The final temperature is \(7\,^\circ\text{C}\). 3. Subtract the initial temperature from the final temperature: \(7-(-12)=7+12=19\).

Answer

The temperature increased by \(19\,^\circ\text{C}\).
5121707
Ms. Weber's bank account balance is \(-\$75.20\). She wants the balance to be exactly \(\$120.00\). How much money must she deposit?

Hints

- A negative balance is below \(0\). - First think about how much is needed to reach \(0\), then how much more is needed to reach \(120\). - Subtract the current signed balance from the target balance.

Solution

1. The current balance is \(-\$75.20\), and the target balance is \(\$120.00\). 2. The required deposit is the change from \(-75.20\) to \(120.00\): \(120.00-(-75.20)=195.20\). 3. Therefore, she must deposit \(\$195.20\).

Answer

\(\$195.20\)
5121737
Ms. Schneider's checking account balance is \(-\$125.50\). In the morning, she deposits \(\$350.00\). In the afternoon, she pays an \(\$80.00\) bill from the account. Find the end-of-day balance using one expression with rational numbers.

Hints

- A deposit increases the account balance. - A bill payment decreases the account balance. - Write all three amounts in one signed-number expression before calculating.

Solution

1. Represent the transactions in one expression: \(-125.50+350.00-80.00\). 2. Calculate: \(-125.50+350.00=224.50\), then \(224.50-80.00=144.50\). 3. The end-of-day balance is \(\$144.50\).

Answer

\(-125.50+350.00-80.00=144.50\). The end-of-day balance is \(\$144.50\).
5121767
Evaluate \(T(a,b)=a-b\) for each pair of rational numbers. Complete the table using fractions in simplest form or mixed numbers. <table> <tr><td>\(a-b\)</td><td>\(-\frac{2}{3}\)</td><td>\(\frac{5}{6}\)</td></tr> <tr><td>\(-\frac{1}{2}\)</td><td></td><td></td></tr> <tr><td>\(\frac{3}{4}\)</td><td></td><td></td></tr> </table>

Hints

- Use a common denominator before subtracting fractions. - Subtracting a negative number means adding. - Simplify each result or write it as a mixed number.

Solution

1. \(-\frac{1}{2}-\left(-\frac{2}{3}\right)=-\frac{3}{6}+\frac{4}{6}=\frac{1}{6}\). 2. \(-\frac{1}{2}-\frac{5}{6}=-\frac{3}{6}-\frac{5}{6}=-\frac{8}{6}=-\frac{4}{3}=-1\frac{1}{3}\). 3. \(\frac{3}{4}-\left(-\frac{2}{3}\right)=\frac{9}{12}+\frac{8}{12}=\frac{17}{12}=1\frac{5}{12}\). 4. \(\frac{3}{4}-\frac{5}{6}=\frac{9}{12}-\frac{10}{12}=-\frac{1}{12}\).

Answer

<table> <tr><td>\(a-b\)</td><td>\(-\frac{2}{3}\)</td><td>\(\frac{5}{6}\)</td></tr> <tr><td>\(-\frac{1}{2}\)</td><td>\(\frac{1}{6}\)</td><td>\(-1\frac{1}{3}\)</td></tr> <tr><td>\(\frac{3}{4}\)</td><td>\(1\frac{5}{12}\)</td><td>\(-\frac{1}{12}\)</td></tr> </table>
5121787
Determine whether each result is positive or negative without finding its exact value. For each part, give a sign rule or an absolute-value comparison that proves your choice; an exact numerical result alone is not a justification. a) \(-12+(-15)\) b) \(24-(-10)\) c) \(-0.5+0.3\) d) \(\frac{1}{2}-\frac{3}{4}\) e) \(-8-(-5)\)

Hints

- Base each decision on signs, opposites, or relative magnitudes rather than exact arithmetic. - For unlike signs, compare absolute values. - Rewrite subtraction of a negative before deciding its effect.

Solution

1. a) The sum of two negative numbers is negative. 2. b) Subtracting a negative is adding a positive, so the result is positive. 3. c) The addends have unlike signs and \(0.5>0.3\), so the sign is negative. 4. d) Since \(\frac{3}{4}>\frac{1}{2}\), subtracting the larger positive number gives a negative result. 5. e) Rewrite as \(-8+5\). Since \(8>5\), the result is negative.

Answer

a) negative — both addends are negative b) positive — subtracting a negative adds a positive c) negative — \(|-0.5|>|0.3|\) d) negative — \(\frac{3}{4}>\frac{1}{2}\) e) negative — \(-8-(-5)=-8+5\) and \(8>5\)
5122367
Rewrite the expression as an equivalent sum that groups compatible fractions, then evaluate it. \(\frac{7}{15} - \frac{3}{4} + \frac{8}{15} - \frac{1}{4}\)

Hints

- First view every subtraction as addition of a signed term. - Which terms can be combined without creating a new denominator? - Include the regrouped expression in your response.

Solution

1. Rewrite and regroup: \(\left(\frac{7}{15}+\frac{8}{15}\right)+\left(-\frac{3}{4}-\frac{1}{4}\right)\). 2. The groups are \(1\) and \(-1\). 3. Their sum is \(0\).

Answer

\(\left(\frac{7}{15}+\frac{8}{15}\right)+\left(-\frac{3}{4}-\frac{1}{4}\right)=0\)
5122377
Evaluate without converting every number to the same form. Show an equivalent grouping that lets the decimals combine with decimals and the fractions combine with fractions. \(2.7 + \frac{2}{9} - 1.7 + \frac{16}{9}\)

Hints

- Keep each sign attached to its term when changing order. - Look for same-form terms that combine exactly as written. - Do not convert a repeating fraction to a decimal just to make every term look alike.

Solution

1. Regroup as \((2.7-1.7)+\left(\frac{2}{9}+\frac{16}{9}\right)\). 2. The groups equal \(1\) and \(2\). 3. Therefore, the value is \(3\).

Answer

\((2.7-1.7)+\left(\frac{2}{9}+\frac{16}{9}\right)=1+2=3\)
5122387
For each expression, write a regrouped equivalent expression that creates convenient pairs, then evaluate. a) \(12.3 - 4.5 - 2.3 + 1.5\) b) \(\frac{3}{8} - 1.4 + \frac{5}{8} - 0.6\)

Hints

- Rewrite each subtraction as addition of a signed term before changing order. - Look for pairs whose sum or difference is a whole number. - Show the regrouping explicitly.

Solution

1. a) Regroup as \((12.3-2.3)+(-4.5+1.5)=10-3=7\). 2. b) Regroup as \(\left(\frac{3}{8}+\frac{5}{8}\right)+(-1.4-0.6)=1-2=-1\).

Answer

a) \((12.3-2.3)+(-4.5+1.5)=7\) b) \(\left(\frac{3}{8}+\frac{5}{8}\right)+(-1.4-0.6)=-1\)
5122417
For each expression, rearrange and group terms to create at least one multiple of \(10\) or \(100\). Write the regrouped expression, then evaluate. a) \(34 + 128 - 134\) b) \(12.5 - 6.7 - 2.5 + 1.7\) c) \(-48 + 56 - 52 + 44\)

Hints

- Keep each sign attached to its term when rearranging. - Search for pairs whose combined value has a zero in the ones place. - Your response must display the convenient grouping.

Solution

1. a) \((34-134)+128=-100+128=28\). 2. b) \((12.5-2.5)+(-6.7+1.7)=10-5=5\). 3. c) \((-48-52)+(56+44)=-100+100=0\).

Answer

a) \((34-134)+128=28\) b) \((12.5-2.5)+(-6.7+1.7)=5\) c) \((-48-52)+(56+44)=0\)
5122447
For each expression, show one regrouping that creates a whole-number partial sum or difference, then evaluate. a) \(15.6 - 7.8 - 2.2\) b) \(-4.9 + 12.5 - 5.1\) c) \(24.3 + 8.7 - 14.3 + 1.3\)

Hints

- Look for decimal pairs whose sum or difference is a whole number. - Preserve the sign of every term when changing grouping. - Include the regrouped expression in your answer.

Solution

1. a) \(15.6-(7.8+2.2)=15.6-10=5.6\). 2. b) \((-4.9-5.1)+12.5=-10+12.5=2.5\). 3. c) \((24.3-14.3)+(8.7+1.3)=10+10=20\).

Answer

a) \(15.6-(7.8+2.2)=5.6\) b) \((-4.9-5.1)+12.5=2.5\) c) \((24.3-14.3)+(8.7+1.3)=20\)
5122477
Use properties of addition to evaluate efficiently. \(7.25-4.8+\frac{11}{4}-1.2\)

Hints

- Convert the fraction to a decimal. - Look for values whose decimal parts combine to whole numbers. - Rewrite the expression as a sum before reordering terms.

Solution

1. Write \(\frac{11}{4}=2.75\). 2. Rewrite subtraction as addition and regroup: \((7.25+2.75)+(-4.8-1.2)\). 3. Evaluate the grouped sums: \(10+(-6)=4\).

Answer

\(4\)
5122497
Rearrange and group the terms to create convenient multiples of \(100\). Write the regrouped expression, then evaluate. \(-250 + 68 + 145 - 50 + 32 - 45\)

Hints

- Keep the sign attached to each term when rearranging. - Look for pairs that combine to exact hundreds. - Show the regrouping before the final total.

Solution

1. Regroup as \((-250-50)+(68+32)+(145-45)\). 2. The groups are \(-300\), \(100\), and \(100\). 3. The total is \(-100\).

Answer

\((-250-50)+(68+32)+(145-45)=-300+100+100=-100\)
5122837
For each pair, write and evaluate an absolute-difference expression for the distance between the numbers. Then decide which pair is farther apart. Pair A: \(-12\) and \(8\) Pair B: \(-5.5\) and \(-20\)

Hints

- Use the same absolute-difference structure for both pairs. - The signed difference may depend on subtraction order, but its absolute value does not. - Compare the two nonnegative distances after evaluating them.

Solution

1. Pair A: \(|8-(-12)|=|20|=20\). 2. Pair B: \(|-5.5-(-20)|=|14.5|=14.5\). 3. Since \(20>14.5\), Pair A is farther apart.

Answer

Pair A: \(|8-(-12)|=20\) Pair B: \(|-5.5-(-20)|=14.5\) Pair A is farther apart.
5123047
For each expression, write one equivalent regrouping that produces convenient whole-number partial results, then evaluate. a) \(14.7 - 3.9 + 5.3 - 6.1\) b) \(-\frac{3}{8} + 12.5 - 5.5 + 0.375\) c) \(8.4 - (2.4 + 3.9) + 0.9\)

Hints

- Search for pairs that make whole numbers or additive inverses. - A fraction-decimal conversion is useful only if it creates such a pair. - Keep signs attached to terms when regrouping.

Solution

1. a) \((14.7+5.3)+(-3.9-6.1)=20-10=10\). 2. b) Since \(-\frac{3}{8}=-0.375\), \((-0.375+0.375)+(12.5-5.5)=7\). 3. c) Remove the parentheses and regroup: \((8.4-2.4)+(0.9-3.9)=6-3=3\).

Answer

a) \((14.7+5.3)+(-3.9-6.1)=10\) b) \((-0.375+0.375)+(12.5-5.5)=7\) c) \((8.4-2.4)+(0.9-3.9)=3\)
5123077
Evaluate the expression: \(\frac{2}{3}-\left(-\frac{1}{4}\right)+\frac{5}{6}-1\)

Hints

- What happens when you subtract a negative number? - Use a common denominator for all fractions. - Write \(1\) as a fraction with the same denominator.

Solution

1. Subtracting a negative gives addition: \(\frac{2}{3}+\frac{1}{4}+\frac{5}{6}-1\). 2. Use denominator \(12\): \(\frac{8}{12}+\frac{3}{12}+\frac{10}{12}-\frac{12}{12}=\frac{9}{12}\). 3. Simplify: \(\frac{9}{12}=\frac{3}{4}=0.75\).

Answer

\(\frac{3}{4}\), or \(0.75\)
5128337
Evaluate each expression. Use common denominators where needed. a) \(\frac{4}{5}+\left(-\frac{1}{2}\right)\) b) \(-\frac{3}{4}-\frac{5}{6}\) c) \(-1.5+\frac{1}{4}\) d) \(\frac{7}{9}-\left(-\frac{2}{3}\right)\)

Hints

- Subtracting a negative number is the same as adding its opposite. - Fractions need common denominators before addition or subtraction. - Convert between fractions and decimals when helpful. - Simplify or write improper fractions as mixed numbers if useful.

Solution

1. For a), \(\frac{4}{5}-\frac{1}{2}=\frac{8}{10}-\frac{5}{10}=\frac{3}{10}\). 2. For b), \(-\frac{3}{4}-\frac{5}{6}=-\frac{9}{12}-\frac{10}{12}=-\frac{19}{12}=-1 \frac{7}{12}\). 3. For c), \(-1.5+0.25=-1.25=-\frac{5}{4}\). 4. For d), subtracting a negative gives addition: \(\frac{7}{9}+\frac{2}{3}=\frac{7}{9}+\frac{6}{9}=\frac{13}{9}=1 \frac{4}{9}\).

Answer

a) \(\frac{3}{10}\) b) \(-\frac{19}{12}\), or \(-1 \frac{7}{12}\) c) \(-1.25\), or \(-\frac{5}{4}\) d) \(\frac{13}{9}\), or \(1 \frac{4}{9}\)
5178397
Complete each statement about how a subtraction difference changes. a) If the minuend increases by \(14\) and the subtrahend stays the same, the difference ______ by ______. b) If the minuend stays the same and the subtrahend decreases by \(9\), the difference ______ by ______. c) If both the minuend and the subtrahend increase by \(20\), then ______.

Hints

- Test each statement with a simple example such as \(20 - 10 = 10\). - What happens to the result when you subtract less? - Think about shifting both numbers the same distance on a number line.

Solution

1. Increasing the minuend by \(14\) while keeping the subtrahend fixed increases the difference by \(14\). 2. Decreasing the subtrahend by \(9\) means \(9\) less is subtracted, so the difference increases by \(9\). 3. Increasing both numbers by the same amount does not change their difference: \((m + 20) - (s + 20) = m - s\).

Answer

a) increases; \(14\) b) increases; \(9\) c) the difference stays the same
5178407
A subtraction expression is written as \(m - s = d\). Describe how \(d\) changes when both changes are made. a) Increase \(m\) by \(30\) and increase \(s\) by \(10\). b) Decrease \(m\) by \(15\) and increase \(s\) by \(15\). c) Increase \(m\) by \(12\) and decrease \(s\) by \(12\).

Hints

- Analyze the change to the minuend and the change to the subtrahend separately. - Increasing the subtrahend decreases the difference; decreasing it increases the difference. - Combine the two effects to find the net change.

Solution

1. Rewrite the expression: \((m + 30) - (s + 10) = m - s + 20\). The difference increases by \(20\). 2. Rewrite the expression: \((m - 15) - (s + 15) = m - s - 30\). The difference decreases by \(30\). 3. Rewrite the expression: \((m + 12) - (s - 12) = m - s + 24\). The difference increases by \(24\).

Answer

a) The difference increases by \(20\). b) The difference decreases by \(30\). c) The difference increases by \(24\).
5178457
Start with \(300 - 120 = 180\). Decrease the minuend by \(50\) and decrease the subtrahend by \(30\). Use how these changes affect the difference to find the new difference. Then state how much it differs from the original difference.

Hints

- Consider the effect of each change separately. - Decreasing the minuend lowers the difference, while decreasing the subtrahend raises it. - Combine the two effects, then apply the net change to the original difference.

Solution

1. Decreasing the minuend by \(50\) decreases the difference by \(50\). 2. Decreasing the subtrahend by \(30\) increases the difference by \(30\). 3. The net change is \(-50 + 30 = -20\), so the new difference is \(180 - 20 = 160\). 4. The new difference is \(20\) less than the original difference.

Answer

The new difference is \(160\). It is \(20\) less than the original difference.
5179097
A difference has a value of \(450\). The minuend is increased by \(120\), and the subtrahend is increased by \(50\). By how much does the difference change, and what is its new value?

Hints

- Analyze the two changes separately. - Increasing the subtrahend means more is subtracted. - Combine the two effects, then apply the net change to \(450\).

Solution

1. Increasing the minuend by \(120\) increases the difference by \(120\). 2. Increasing the subtrahend by \(50\) decreases the difference by \(50\). 3. The net change is \(120 - 50 = 70\), so the difference increases by \(70\). 4. The new value is \(450 + 70 = 520\).

Answer

The difference increases by \(70\), and its new value is \(520\).
5180687
Ms. Meyer has \(\$150\) in her checking account. A debit of \(\$215\) is then posted for a new desk. Explain why the new balance must be negative, and calculate the balance.

Hints

- What happens when more money is withdrawn than the account contains? - Compare the two amounts. Which is greater? - How can you represent the debit when calculating the new balance?

Solution

1. The debit is greater than the current balance because \(215>150\), so the account becomes overdrawn and the new balance is negative. 2. Subtract the debit from the balance: \(150-215=-65\).

Answer

The new balance is negative because the debit is greater than the amount in the account. \(\$150-\$215=-\$65\), so the new balance is \(-\$65\).
5180697
At \(6{:}00\) p.m. on a winter day, the temperature is \(4\,^\circ\text{F}\). By midnight, it drops \(7\,^\circ\text{F}\). During the early morning, it drops another \(2\,^\circ\text{F}\). What is the temperature in the morning?

Hints

- Represent each change on a number line. - Which direction do you move when the temperature drops? - Calculate the first drop, then use that result as the starting value for the second drop.

Solution

1. Find the temperature at midnight: \(4-7=-3\), so it is \(-3\,^\circ\text{F}\). 2. Subtract the second drop: \(-3-2=-5\), so it is \(-5\,^\circ\text{F}\).

Answer

The temperature in the morning is \(-5\,^\circ\text{F}\).
5180757
For each sum, give a sign prediction and a brief rule or magnitude comparison that supports it. Then calculate the exact sum. a) \(-14+25\) b) \(18+(-30)\) c) \(-22+(-18)\) d) \(-45+45\)

Hints

- Separate the sign decision from the magnitude calculation. - With unlike signs, compare absolute values. - Opposites have equal magnitude and opposite signs.

Solution

1. a) The signs differ and \(25>14\), so predict positive; the sum is \(11\). 2. b) The signs differ and \(30>18\), so predict negative; the sum is \(-12\). 3. c) Both addends are negative, so predict negative; the sum is \(-40\). 4. d) The addends are opposites, so predict zero; the sum is \(0\).

Answer

a) positive because \(25>14\); \(11\) b) negative because \(30>18\); \(-12\) c) negative because both addends are negative; \(-40\) d) zero because the addends are opposites; \(0\)
5180767
Find the integer that makes each equation true. a) \(-12+\square=5\) b) \(\square+(-8)=-20\) c) \(15+\square=0\) d) \(\square+10=-3\)

Hints

- Think of moving from the known addend to the sum on a number line. - Use subtraction to find a missing addend. - A number and its opposite add to zero.

Solution

1. In part a), the missing addend is \(5-(-12)=17\). 2. In part b), the missing addend is \(-20-(-8)=-12\). 3. In part c), the missing addend is the opposite of \(15\), so it is \(-15\). 4. In part d), the missing addend is \(-3-10=-13\).

Answer

a) \(17\) b) \(-12\) c) \(-15\) d) \(-13\)
5180777
Replace each box with \(<\), \(>\), or \(=\). For each part, give the sign rule or magnitude comparison that lets you decide without finding the exact sum. a) \(-35+40\mathbin{\square}0\) b) \(-12+(-15)\mathbin{\square}0\) c) \(25+(-25)\mathbin{\square}0\) d) \(-50+30\mathbin{\square}0\)

Hints

- Do not compute the sums; use signs and absolute values. - For unlike signs, the larger magnitude determines the sign. - Equal opposite magnitudes produce zero.

Solution

1. a) The positive addend has greater magnitude, so the sum is greater than \(0\). 2. b) Both addends are negative, so the sum is less than \(0\). 3. c) The addends are opposites, so the sum equals \(0\). 4. d) The negative addend has greater magnitude, so the sum is less than \(0\).

Answer

a) \(>\) — \(40>35\) b) \(<\) — both addends are negative c) \(=\) — the addends are opposites d) \(<\) — \(50>30\) and the larger magnitude is negative
5180857
Find each sum. a) \(-38+92\) b) \(54+(-117)\) c) \(-256+(-144)\) d) \(-1025+375\)

Hints

- First identify whether the signs are alike or different. - With unlike signs, compare absolute values. - With two negative addends, add the absolute values and keep the negative sign.

Solution

1. For unlike signs, subtract absolute values and use the sign of the addend with the greater absolute value. 2. \(-38+92=54\). 3. \(54+(-117)=-(117-54)=-63\). 4. \(-256+(-144)=-(256+144)=-400\). 5. \(-1025+375=-(1025-375)=-650\).

Answer

a) \(54\) b) \(-63\) c) \(-400\) d) \(-650\)
5180867
Evaluate \(-120+450+(-380)\) from left to right. Report the intermediate sum after the first addition and the final sum.

Hints

- The stated order requires you to combine the first two terms first. - Keep the sign of the third term when adding it to the intermediate result.

Solution

1. First, \(-120+450=330\). 2. Then \(330+(-380)=-50\).

Answer

Intermediate sum: \(330\) Final sum: \(-50\)
5180877
Compare the values of Expressions A and B. Which is greater? Expression A: \(-1500+850\) Expression B: \(2100+(-2800)\)

Hints

- Evaluate each expression separately. - Compare the two negative results on a number line. - Among negative numbers, the value closer to zero is greater.

Solution

1. Expression A is \(-1500+850=-650\). 2. Expression B is \(2100+(-2800)=-700\). 3. Since \(-650>-700\), Expression A is greater.

Answer

Expression A is greater because \(-650>-700\).
5180907
A research submarine is \(450\,\text{m}\) below sea level. It rises \(185\,\text{m}\) to take a measurement. How far below sea level is the submarine after it rises?

Hints

- Represent positions below sea level with negative numbers. - When the submarine rises, does its position become greater or less? - Express the final negative position as a depth below sea level.

Solution

1. Represent the initial position as \(-450\,\text{m}\). 2. Rising \(185\,\text{m}\) means adding \(185\): \(-450+185=-265\). 3. A position of \(-265\,\text{m}\) is \(265\,\text{m}\) below sea level.

Answer

The submarine is \(265\,\text{m}\) below sea level.
5181017
Choose \(+\) or \(-\) for each box to make the equation true. a) \((\mathbin{\square}17)+13=30\) b) \(-24+(\mathbin{\square}16)=-40\) c) \(35+(\mathbin{\square}50)=-15\) d) \((\mathbin{\square}12)+(-28)=-16\)

Hints

- Check whether the result must increase or decrease from the known addend. - Compare the absolute values when the signs differ. - Substitute each possible sign and test the equation.

Solution

1. \(17+13=30\), so part a) needs \(+\). 2. \(-24+(-16)=-40\), so part b) needs \(-\). 3. \(35+(-50)=-15\), so part c) needs \(-\). 4. \(12+(-28)=-16\), so part d) needs \(+\).

Answer

a) \(+\) b) \(-\) c) \(-\) d) \(+\)
5181027
Choose the missing sign in each equation. a) \(-42+12=\mathbin{\square}30\) b) \(15+(-45)=\mathbin{\square}30\) c) \((\mathbin{\square}60)+(-25)=35\) d) \(-11+(\mathbin{\square}11)=-22\)

Hints

- Identify whether the missing sign belongs to an addend or the result. - For unlike signs, subtract the absolute values. - Test the completed equation.

Solution

1. \(-42+12=-30\), so part a) needs \(-\). 2. \(15+(-45)=-30\), so part b) needs \(-\). 3. \(60+(-25)=35\), so part c) needs \(+\). 4. \(-11+(-11)=-22\), so part d) needs \(-\).

Answer

a) \(-\) b) \(-\) c) \(+\) d) \(-\)
5181047
For each sum, state the predicted sign and the rule or magnitude comparison that determines it. Then calculate the exact value. a) \(-415+625\) b) \(-874+(-126)\) c) \(2500+(-3750)\) d) \(-999+1001\)

Hints

- Decide the sign using sign rules and absolute values before reporting the exact sum. - With unlike signs, compare magnitudes. - With like negative signs, the sum remains negative.

Solution

1. a) Unlike signs with \(625>415\) give a positive result: \(210\). 2. b) Two negative addends give a negative result: \(-1000\). 3. c) Unlike signs with \(3750>2500\) give a negative result: \(-1250\). 4. d) Unlike signs with \(1001>999\) give a positive result: \(2\).

Answer

a) positive because \(625>415\); \(210\) b) negative because both addends are negative; \(-1000\) c) negative because \(3750>2500\); \(-1250\) d) positive because \(1001>999\); \(2\)
5181057
Evaluate each sum from left to right. For each part, report the intermediate sum after the first addition and then the final sum. a) \(-120+250+(-80)\) b) \(500+(-750)+250\)

Hints

- Follow the stated order rather than regrouping. - Keep the intermediate value visible before using the final term.

Solution

1. a) \(-120+250=130\), then \(130-80=50\). 2. b) \(500-750=-250\), then \(-250+250=0\).

Answer

a) Intermediate: \(130\); final: \(50\) b) Intermediate: \(-250\); final: \(0\)
5181167
Evaluate both sides and replace each box with \(<\), \(>\), or \(=\). a) \(-45+20\mathbin{\square}-30+5\) b) \(100+(-150)\mathbin{\square}-80+20\) c) \(-12+(-13)\mathbin{\square}5+(-30)\)

Hints

- Evaluate the left and right expressions separately. - Compare negative results carefully. - Among two negative values, the one closer to zero is greater.

Solution

1. In part a), both sides equal \(-25\), so use \(=\). 2. In part b), the left side is \(-50\) and the right side is \(-60\), so use \(>\). 3. In part c), both sides equal \(-25\), so use \(=\).

Answer

a) \(=\) b) \(>\) c) \(=\)
5181177
Find the integer \(z\) that makes each equation true. a) \(50+z=-10\) b) \(z+(-120)=-200\) c) \(-15+25+z=0\)

Hints

- Use subtraction to find a missing addend. - Simplify known addends first. - A number and its opposite sum to zero.

Solution

1. In part a), \(z=-10-50=-60\). 2. In part b), \(z=-200-(-120)=-80\). 3. In part c), \(-15+25=10\), so \(10+z=0\) and \(z=-10\).

Answer

a) \(z=-60\) b) \(z=-80\) c) \(z=-10\)
5181207
A frozen-food warehouse has several storage zones. Zone A is kept at \(-18\,^\circ\text{C}\). Zone B is \(7\) degrees colder than Zone A. Zone C is \(12\) degrees warmer than Zone B. What are the temperatures in Zones B and C?

Hints

- Find the temperature in Zone B first. - How do “colder” and “warmer” translate into operations? - Check the sign of your intermediate result before completing the second calculation.

Solution

1. Zone B is \(7\) degrees colder than Zone A, so \(-18-7=-25\). Zone B is \(-25\,^\circ\text{C}\). 2. Zone C is \(12\) degrees warmer than Zone B, so \(-25+12=-13\). Zone C is \(-13\,^\circ\text{C}\).

Answer

Zone B is \(-25\,^\circ\text{C}\), and Zone C is \(-13\,^\circ\text{C}\).
5181257
Insert \(+\) or \(-\) in each circle to make the equation true. a) \((\bigcirc 75)+(-25)=-100\) b) \((\bigcirc 120)+(+80)=-40\) c) \((+15)+(\bigcirc 50)=+65\) d) \((\bigcirc 9)+(+21)=+12\)

Hints

- Test whether each unknown number must be positive or negative. - Think about movement to the left or right on a number line. - Adding a negative number moves the value left.

Solution

1. In a), \(-75+(-25)=-100\), so the sign is \(-\). 2. In b), \(-120+80=-40\), so the sign is \(-\). 3. In c), \(15+50=65\), so the sign is \(+\). 4. In d), \(-9+21=12\), so the sign is \(-\).

Answer

a) \(-\) b) \(-\) c) \(+\) d) \(-\)
5181347
Find the missing addend in each equation. a) \(15+\square=8\) b) \(15+\square=-2\) c) \(15+\square=0\) d) \(15+\square=16\)

Hints

- Decide whether the missing addend moves the value left or right on a number line. - Use subtraction to find a missing addend. - Substitute your answer to check each equation.

Solution

1. In each equation \(15+x=s\), calculate \(x=s-15\). 2. The results are \(8-15=-7\), \(-2-15=-17\), \(0-15=-15\), and \(16-15=1\).

Answer

a) \(-7\) b) \(-17\) c) \(-15\) d) \(1\)
5181357
For each equation \(a+x=s\), find \(x\). a) \(-7+x=-15\) b) \(-7+x=4\) c) \(-7+x=-7\) d) \(-7+x=20\)

Hints

- Find a missing addend by subtracting the known addend from the sum. - Subtracting \(-7\) is the same as adding \(7\). - Check each result in the original equation.

Solution

1. Use \(x=s-a=s-(-7)=s+7\). 2. The values are \(-15+7=-8\), \(4+7=11\), \(-7+7=0\), and \(20+7=27\).

Answer

a) \(-8\) b) \(11\) c) \(0\) d) \(27\)
5181367
Find the missing addend. a) What number added to \(10\) gives \(-5\)? b) What number added to \(-20\) gives \(0\)? c) What number added to \(-4\) gives \(-12\)? d) What number added to \(0\) gives \(-8\)?

Hints

- Write an addition equation for each question. - Use subtraction to find the unknown addend. - Opposites add to zero.

Solution

1. \(10+x=-5\), so \(x=-15\). 2. \(-20+x=0\), so \(x=20\). 3. \(-4+x=-12\), so \(x=-8\). 4. \(0+x=-8\), so \(x=-8\).

Answer

a) \(-15\) b) \(20\) c) \(-8\) d) \(-8\)
5181507
Evaluate \(154+(-289)+45\).

Hints

- Combine the positive addends first. - Compare the absolute values of the remaining terms. - The term with the greater absolute value determines the sign.

Solution

1. Combine the positive addends: \(154+45=199\). 2. Then \(199+(-289)=-(289-199)=-90\).

Answer

\(-90\)
5181517
Evaluate \(-72+[148+(-205)]\).

Hints

- Evaluate the bracketed expression first. - Use the rule for adding integers with unlike signs. - Then add the two negative values.

Solution

1. Evaluate inside the brackets: \(148+(-205)=-57\). 2. Then \(-72+(-57)=-(72+57)=-129\).

Answer

\(-129\)
5181527
Evaluate \(-120+55+(-80)+145\).

Hints

- Regroup addends to make convenient sums. - Combine negative and positive terms separately. - Opposites add to zero.

Solution

1. Group like-signed terms: \([-120+(-80)]+[55+145]\). 2. The negative terms total \(-200\), and the positive terms total \(200\). 3. Therefore, \(-200+200=0\).

Answer

\(0\)
5181697
Which choice is closest to the value of \(-748+102+(-251)\)? A) \(-1100\) B) \(-900\) C) \(-400\) D) \(-650\)

Hints

- Round to convenient hundreds or fifties. - Combine the negative terms first. - Compare your estimate with the choices.

Solution

1. Round the terms to \(-750\), \(100\), and \(-250\). 2. The estimated sum is \(-750+100-250=-900\). 3. The exact value is \(-897\), confirming that \(-900\) is closest.

Answer

B) \(-900\)
5181787
Decide whether each statement is true or false. Justify with an example or counterexample. a) The sum of two negative numbers is always positive. b) Adding the opposite of \(0\) to any number leaves the number unchanged.

Hints

- Test the first claim with two specific negative integers. - Find the opposite of zero. - One counterexample disproves an “always” claim.

Solution

1. Statement a) is false. For example, \(-3+(-5)=-8\), which is negative. 2. Statement b) is true. The opposite of \(0\) is \(0\), and \(a+0=a\) for every number \(a\).

Answer

a) False; for example, \(-3+(-5)=-8\). b) True; the opposite of \(0\) is \(0\), and adding \(0\) does not change a number.
5181837
Find the missing addend. a) \(-520+\square=180\) b) \(75+\square=-25\)

Hints

- Decide whether the movement is left or right on a number line. - Determine whether the move crosses zero. - Subtract the starting value from the target value.

Solution

1. In part a), the move from \(-520\) to \(180\) is \(520+180=700\), so the missing addend is \(700\). 2. In part b), the move from \(75\) to \(-25\) is \(-(75+25)=-100\), so the missing addend is \(-100\).

Answer

a) \(700\) b) \(-100\)
5181847
Find \(x\). a) \(x+(-150)=-200\) b) \(x+300=50\)

Hints

- Undo the added number to find the starting value. - Adding a negative number can be undone by adding its opposite. - Check each solution in the original equation.

Solution

1. In part a), add \(150\) to both sides: \(x=-200+150=-50\). 2. In part b), subtract \(300\) from both sides: \(x=50-300=-250\).

Answer

a) \(x=-50\) b) \(x=-250\)
5181857
Find \(x\). a) \(-12+x=-12\) b) \(44+x=0\)

Hints

- Which addend leaves a number unchanged? - Which addend makes a sum of zero? - Use the additive identity and additive inverse.

Solution

1. In part a), adding \(0\) leaves \(-12\) unchanged, so \(x=0\). 2. In part b), the number that adds to \(44\) to make zero is its opposite, so \(x=-44\).

Answer

a) \(x=0\) b) \(x=-44\)
5181887
For each pair, find \(s=a+b\), then decide whether \(s>a\). a) \(a=12\), \(b=-5\) b) \(a=-6\), \(b=-4\) c) \(a=-9\), \(b=10\)

Hints

- Calculate each sum first. - Compare the result with the first addend. - Consider how a positive or negative second addend changes the first.

Solution

1. a) \(12+(-5)=7\), and \(7>12\) is false. 2. b) \(-6+(-4)=-10\), and \(-10>-6\) is false. 3. c) \(-9+10=1\), and \(1>-9\) is true.

Answer

a) \(s=7\); no b) \(s=-10\); no c) \(s=1\); yes
5182167
On a winter afternoon, the temperature drops by \(9\,^\circ\text{F}\). That evening, the thermometer reads \(-5\,^\circ\text{F}\). What was the temperature before the drop?

Hints

- Was it warmer or colder before the temperature dropped? - Sketch the situation on a number line or thermometer. - Starting with the final value, use the inverse of the temperature change.

Solution

1. Let the initial temperature be the value that becomes \(-5\,^\circ\text{F}\) after a decrease of \(9\) degrees. 2. Undo the decrease by adding \(9\): \(-5+9=4\), so the initial temperature was \(4\,^\circ\text{F}\).

Answer

The temperature before the drop was \(4\,^\circ\text{F}\).
5182177
Mr. Weber checks his banking app. A credit of \(\$125\) has been added, and his current balance is \(-\$240\). What was his balance immediately before the credit?

Hints

- What effect does a credit have on an account balance? - Which inverse operation undoes the credit? - Remember that the current balance is already negative.

Solution

1. The credit increased the account balance by \(\$125\). 2. Undo the credit by subtracting \(125\) from the current balance: \(-240-125=-365\).

Answer

The balance before the credit was \(-\$365\).
5182187
A remotely operated underwater vehicle descends \(150\,\text{m}\) from its initial position. It ends at \(-620\,\text{m}\), where sea level is \(0\,\text{m}\). What was the vehicle's position before it descended?

Hints

- Treat sea level as \(0\); positions below it are negative. - Descending makes the position less, or more negative. - To find the earlier position, reverse the descent.

Solution

1. The final position is \(-620\,\text{m}\). 2. Undo the \(150\,\text{m}\) descent by adding \(150\): \(-620+150=-470\). 3. Check: \(-470-150=-620\).

Answer

The vehicle's initial position was \(-470\,\text{m}\), or \(470\,\text{m}\) below sea level.
5182277
Rewrite each subtraction as addition of the opposite, and then evaluate. a) \(45-17\) b) \(-32-(-58)\) c) \(12-(-88)\) d) \(-74-26\)

Hints

- Subtracting a number means adding its opposite. - Change the operation and the sign of the second number together. - Then use integer addition rules.

Solution

1. \(45-17=45+(-17)=28\). 2. \(-32-(-58)=-32+58=26\). 3. \(12-(-88)=12+88=100\). 4. \(-74-26=-74+(-26)=-100\).

Answer

a) \(45+(-17)=28\) b) \(-32+58=26\) c) \(12+88=100\) d) \(-74+(-26)=-100\)
5182287
Complete each equation by rewriting subtraction as addition of the opposite. a) \(-120-85=-120+(\square)=\square\) b) \(215-(-45)=215+(\square)=\square\) c) \(-63-(-163)=-63+(\square)=\square\) d) \(19-50=19+(\square)=\square\)

Hints

- Replace subtraction with addition. - Use the opposite of the number after the subtraction sign. - Evaluate the resulting sum.

Solution

1. \(-120-85=-120+(-85)=-205\). 2. \(215-(-45)=215+45=260\). 3. \(-63-(-163)=-63+163=100\). 4. \(19-50=19+(-50)=-31\).

Answer

a) \(-120+(-85)=-205\) b) \(215+45=260\) c) \(-63+163=100\) d) \(19+(-50)=-31\)
5182297
Rewrite each expression as a sum, evaluate it, and decide which value is greater. Expression A: \(-240-160\) Expression B: \(-240-(-160)\)

Hints

- Rewrite each subtraction as adding the opposite. - Evaluate both expressions separately. - The negative value closer to zero is greater.

Solution

1. Expression A becomes \(-240+(-160)=-400\). 2. Expression B becomes \(-240+160=-80\). 3. Since \(-80>-400\), Expression B is greater.

Answer

Expression A equals \(-400\). Expression B equals \(-80\). Expression B is greater.
5182327
Order the expressions by value. Which expressions have equal values? 1. \(3-8\) 2. \(-2-3\) 3. \(12-4\) 4. \(-2-(-10)\) 5. \(0-5\) 6. \(15-7\)

Hints

- Rewrite subtraction as adding the opposite. - Evaluate each expression separately. - Group expressions with equal results.

Solution

1. Expressions 1, 2, and 5 each equal \(-5\). 2. Expressions 3, 4, and 6 each equal \(8\). 3. Therefore, \(1=2=5<3=4=6\).

Answer

Order: \(1=2=5<3=4=6\) Value \(-5\): 1, 2, 5 Value \(8\): 3, 4, 6
5182347
Evaluate both sides and replace each box with \(<\), \(>\), or \(=\). a) \(-8-2\mathbin{\square}-8-(-2)\) b) \(5-5\mathbin{\square}-3-(-3)\) c) \(-12-4\mathbin{\square}-10-10\)

Hints

- Evaluate each side separately. - Subtracting a negative is adding a positive. - Compare negative values on a number line.

Solution

1. In a), the values are \(-10\) and \(-6\), so \(-10<-6\). 2. In b), both values are \(0\), so they are equal. 3. In c), the values are \(-16\) and \(-20\), so \(-16>-20\).

Answer

a) \(<\) b) \(=\) c) \(>\)
5182397
Evaluate both sides and replace each box with \(<\), \(>\), or \(=\). a) \(-58-22\mathbin{\square}-90+10\) b) \(72+(-84)\mathbin{\square}-5-(-7)\) c) \(-200+150\mathbin{\square}-10-45\)

Hints

- Evaluate left and right sides separately. - Rewrite subtraction of a negative before calculating. - The negative value closer to zero is greater.

Solution

1. In a), both sides equal \(-80\), so use \(=\). 2. In b), the left side is \(-12\) and the right side is \(2\), so use \(<\). 3. In c), the left side is \(-50\) and the right side is \(-55\), so use \(>\).

Answer

a) \(=\) b) \(<\) c) \(>\)
5182467
Insert \(+\) or \(-\) in each box to make the equation true. a) \((\square14)+(-26)=-40\) b) \((-55)-(\square25)=-30\) c) \((+12)+(\square18)=-6\)

Hints

- Determine whether each unknown number must be positive or negative. - Subtracting a negative number increases the value. - Substitute each possible sign and verify the equation.

Solution

1. In a), \(-14+(-26)=-40\), so the sign is \(-\). 2. In b), \(-55-(-25)=-30\), so the sign is \(-\). 3. In c), \(12+(-18)=-6\), so the sign is \(-\).

Answer

a) \(-\) b) \(-\) c) \(-\)
5182537
A research submarine is \(85\,\text{m}\) below sea level. It descends another \(40\,\text{m}\). Give its new position relative to sea level as an integer.

Hints

- How do you represent a position below sea level with a signed number? - Does descending make the position greater or less? - Is the submarine moving closer to or farther from sea level?

Solution

1. Represent the initial position as \(-85\). 2. A descent of \(40\,\text{m}\) means subtracting \(40\): \(-85-40=-125\).

Answer

The new position is \(-125\), or \(125\,\text{m}\) below sea level.
5182647
Evaluate each difference. First rewrite the subtraction as addition of the opposite. a) \((-410)-(+590)\) b) \((+1200)-(-800)\) c) \((-33)-(-33)\) d) \((-75)-(+25)\)

Hints

- To subtract a number, add its opposite. - Pay attention to what happens when a number is subtracted from itself. - Subtracting a positive number from a negative number makes the result more negative.

Solution

1. \((-410)-(+590)=(-410)+(-590)=-1000\). 2. \((+1200)-(-800)=(+1200)+(+800)=2000\). 3. \((-33)-(-33)=(-33)+(+33)=0\). 4. \((-75)-(+25)=(-75)+(-25)=-100\).

Answer

a) \((-410)+(-590)=-1000\) b) \((+1200)+(+800)=2000\) c) \((-33)+(+33)=0\) d) \((-75)+(-25)=-100\)
5182657
For each value of \(x\), evaluate \((-150)-x\). First rewrite the subtraction as addition. a) \(x=50\) b) \(x=-50\) c) \(x=150\) d) \(x=-200\)

Hints

- Substitute each signed value for \(x\). - To subtract a number, add its opposite. - The result can be positive when a negative number is subtracted.

Solution

1. For \(x=50\), \((-150)-50=(-150)+(-50)=-200\). 2. For \(x=-50\), \((-150)-(-50)=(-150)+50=-100\). 3. For \(x=150\), \((-150)-150=(-150)+(-150)=-300\). 4. For \(x=-200\), \((-150)-(-200)=(-150)+200=50\).

Answer

a) \((-150)+(-50)=-200\) b) \((-150)+50=-100\) c) \((-150)+(-150)=-300\) d) \((-150)+200=50\)
5182747
Ms. Meyer's checking account balance was \(-\$45\) yesterday. Today, two more debits were posted: a \(\$30\) bill and a \(\$25\) shipping charge. What is her current balance?

Hints

- Do debits increase or decrease an account balance? - On a number line, which direction represents a debit? - Apply the two debits one at a time.

Solution

1. After the first debit, the balance is \(-45-30=-75\). 2. After the second debit, the balance is \(-75-25=-100\).

Answer

Her current balance is \(-\$100\).
5182777
Find each value. a) Find the sum of \(-42\) and \(-18\). b) Find the difference of \(-25\) and \(35\), with \(-25\) as the starting number.

Hints

- A sum uses addition, while a difference uses subtraction. - Pay close attention to the signs of both numbers. - A number line can help you check the direction of each operation.

Solution

1. In a), \(-42+(-18)=-60\). 2. In b), \(-25-35=-60\).

Answer

a) \(-60\) b) \(-60\)
5182787
Decide whether each statement is true or false. a) The sum of \(-15\) and \(15\) is \(0\). b) The difference of \(-40\) and \(-10\) is \(-50\). c) The sum of \(-9\) and \(-11\) is \(-20\).

Hints

- Write an expression for each statement. - Subtracting a negative number is the same as adding its opposite. - Compare each calculated value with the value in the statement.

Solution

1. In a), \(-15+15=0\), so the statement is true. 2. In b), \(-40-(-10)=-30\), not \(-50\), so the statement is false. 3. In c), \(-9+(-11)=-20\), so the statement is true.

Answer

a) True b) False c) True
5182817
How much less is \(-245\) than \(-112\)? Write an absolute-difference expression for the distance between the two values, evaluate it, and use that distance to answer the question.

Hints

- Represent the separation of the two signed numbers with an absolute difference. - The distance is nonnegative even though both coordinates are negative. - Use the distance to state how much less one value is than the other.

Solution

1. The distance is \(|-245-(-112)|=|-133|=133\). 2. Therefore, \(-245\) is \(133\) less than \(-112\).

Answer

\(|-245-(-112)|=133\). Therefore, \(-245\) is \(133\) less than \(-112\).
5182827
A number plus \(-67\) equals \(25\). What is the number?

Hints

- Write an equation with a variable. - Undo the added negative number. - Check the result in the original equation.

Solution

1. Write \(x+(-67)=25\). 2. Undo adding \(-67\) by adding \(67\): \(x=25+67=92\).

Answer

\(92\)
5182897
Evaluate \((-45)-(-18)+(-32)\).

Hints

- Subtracting a negative number is the same as adding a positive number. - Work from left to right. - At each step, decide whether the value should increase or decrease.

Solution

1. Rewrite the subtraction: \(-45-(-18)=-45+18=-27\). 2. Then add the last term: \(-27+(-32)=-59\).

Answer

\(-59\)
5183117
Which subtraction expressions have a value of \(-24\)? Write all corresponding letters. A) \(16-40\) B) \(-12-12\) C) \(-30-(-6)\) D) \(0-(-24)\) E) \(-10-14\) F) \(-48-(-24)\)

Hints

- Rewrite subtraction of a negative as addition. - Evaluate each expression separately. - Check whether each result is positive or negative.

Solution

1. A: \(16-40=-24\). 2. B: \(-12-12=-24\). 3. C: \(-30-(-6)=-30+6=-24\). 4. D: \(0-(-24)=24\). 5. E: \(-10-14=-24\). 6. F: \(-48-(-24)=-48+24=-24\).

Answer

A, B, C, E, and F
5183657
First rewrite each expression without parentheses using simplified notation. Then evaluate. a) \((+56)+(-14)\) b) \((-33)-(-17)\) c) \((-120)+(-80)\) d) \((+200)-(+45)\)

Hints

- Distinguish the operation sign from the sign attached to a number. - Adding a negative is subtraction, and subtracting a negative is addition. - A number line can help you check each result.

Solution

1. In a), \(56-14=42\). 2. In b), \(-33+17=-16\). 3. In c), \(-120-80=-200\). 4. In d), \(200-45=155\).

Answer

a) \(56-14=42\) b) \(-33+17=-16\) c) \(-120-80=-200\) d) \(200-45=155\)
5183817
Rewrite the expression as a sum. Then show an equivalent rearrangement or grouping that makes the calculation efficient before evaluating. \(-140+512-(-40)\)

Hints

- Rewrite subtraction of a negative number as addition of its opposite. - Keep the signs attached to the addends when changing their order. - Look for a pair that is simpler to combine before adding the remaining term.

Solution

1. Rewrite subtraction of a negative as addition: \(-140+512+40\). 2. Rearrange and group: \((-140+40)+512\). 3. Evaluate: \(-100+512=412\).

Answer

Sum form: \(-140+512+40\) One valid efficient regrouping: \((-140+40)+512\) Value: \(412\)
5183827
Rewrite the expression as a sum. Then show an equivalent rearrangement or grouping that makes the calculation efficient before evaluating. \(385-119+615\)

Hints

- Rewrite subtraction as addition of the opposite. - Keep each sign attached to its addend when rearranging. - Look for a pair whose sum is especially convenient before using the remaining addend.

Solution

1. Rewrite subtraction as addition of the opposite: \(385+(-119)+615\). 2. Rearrange and group: \((385+615)+(-119)\). 3. Evaluate: \(1000-119=881\).

Answer

Sum form: \(385+(-119)+615\) One valid efficient regrouping: \((385+615)+(-119)\) Value: \(881\)
5183847
Evaluate the expressions. Then group the expressions that have equal values. a) \(24-37\) b) \(37-24\) c) \(-(37-24)\) d) \(-24+37\) e) \(-24-(-37)\) f) \(-37+24\)

Hints

- Evaluate each expression separately. - A negative sign outside parentheses takes the opposite of the parenthetical value. - Subtracting a negative number is the same as adding a positive number.

Solution

1. The values are: a) \(-13\), b) \(13\), c) \(-13\), d) \(13\), e) \(13\), and f) \(-13\). 2. Therefore, a), c), and f) have value \(-13\), while b), d), and e) have value \(13\).

Answer

Value \(-13\): a), c), f) Value \(13\): b), d), e)
5184087
Rewrite without parentheses using simplified notation, and then evaluate: \((-432)-(-218)\).

Hints

- Two consecutive negative signs become addition. - Compare the absolute values of the two numbers. - The number with the greater absolute value determines the sign of the result.

Solution

1. Rewrite the subtraction: \(-432+218\). 2. The difference between the absolute values is \(432-218=214\). 3. Since \(432\) has the greater absolute value and is negative, the result is \(-214\).

Answer

\(-214\)
5184097
Rewrite without parentheses using simplified notation, and then evaluate: \((+675)+(-280)-(+145)\).

Hints

- Rewrite each signed number without parentheses. - You may add the two amounts being subtracted before subtracting their total. - Keep careful track of each operation sign.

Solution

1. Rewrite the expression as \(675-280-145\). 2. Evaluate: \(675-280=395\), and \(395-145=250\).

Answer

\(675-280-145=250\)
5184117
Rewrite each expression as a sum by replacing subtraction with addition of the opposite. Then evaluate. a) \(-215-(-45)-30\) b) \(-128-72+100\)

Hints

- To subtract a number, add its opposite. - The opposite of a positive number is negative, and the opposite of a negative number is positive. - Group addends with the same sign when that makes the calculation easier.

Solution

1. In a), \((-215)+45+(-30)=-170-30=-200\). 2. In b), \((-128)+(-72)+100=-200+100=-100\).

Answer

a) \((-215)+45+(-30)=-200\) b) \((-128)+(-72)+100=-100\)
5184127
Rewrite each expression using addition only, and then evaluate. a) \(440-600-(-160)\) b) \(-19-(-81)-100\)

Hints

- Rewrite each subtraction as addition of the opposite. - A subtraction sign before a positive number becomes addition of a negative number. - Opposites add to zero.

Solution

1. In a), \(440+(-600)+160=-160+160=0\). 2. In b), \((-19)+81+(-100)=62-100=-38\).

Answer

a) \(440+(-600)+160=0\) b) \((-19)+81+(-100)=-38\)
5184357
Evaluate \(48+(52-315)\).

Hints

- Use the order of operations and evaluate the parentheses first. - Adding a negative number is the same as subtracting its absolute value. - Keep track of the sign of the intermediate result.

Solution

1. Evaluate inside the parentheses: \(52-315=-263\). 2. Add the result: \(48+(-263)=-215\).

Answer

\(-215\)
5184987
Lena has four number cards: \(16\), \(8\), \(-8\), and \(24\). She may use addition signs, subtraction signs, and parentheses to form expressions. Each card may be used at most once in an expression. Write two different expressions that each have a value of \(24\).

Hints

- Look for two cards whose values can combine to make \(24\). - Remember that subtracting a negative number is the same as adding a positive number. - The cards \(8\) and \(-8\) can combine to make zero.

Solution

1. One expression is \(16+8\), which equals \(24\). 2. A different expression is \(16-(-8)\), which also equals \(24\). 3. Each number card is used no more than once in either expression.

Answer

Possible answers are \(16+8\) and \(16-(-8)\).
5184997
Which expressions have equal values? Group expressions with the same value, and give that value. A: \(45-(-5)\) B: \(45+5\) C: \(45+(-5)\) D: \(45-5\) E: \(50-0\) F: \(50-10\)

Hints

- Evaluate each expression separately. - Pay close attention to addition or subtraction of a negative number. - Group the letters after comparing all results.

Solution

1. A, B, and E each have value \(50\): \(45-(-5)=50\), \(45+5=50\), and \(50-0=50\). 2. C, D, and F each have value \(40\): \(45+(-5)=40\), \(45-5=40\), and \(50-10=40\).

Answer

Value \(50\): A, B, E Value \(40\): C, D, F
5185387
Evaluate each expression. a) \(-2500+100{,}500\) b) \(-340{,}000-12{,}500\) c) \(50{,}200+(-45{,}200)\) d) \(12{,}300-(-7700)\)

Hints

- Decide whether each result should be positive or negative before calculating. - Adding a negative is subtraction, and subtracting a negative is addition. - Align place values carefully when working with large integers.

Solution

1. In a), \(100{,}500-2500=98{,}000\). 2. In b), \(-340{,}000-12{,}500=-352{,}500\). 3. In c), \(50{,}200-45{,}200=5000\). 4. In d), \(12{,}300+7700=20{,}000\).

Answer

a) \(98{,}000\) b) \(-352{,}500\) c) \(5000\) d) \(20{,}000\)
5186377
Consider the expression \((560 - 140) + (320 + 180)\). Without fully evaluating the new expression, determine how its value changes if each of the four numbers is increased by \(15\). Explain your reasoning from the structure of the expression.

Hints

- Analyze each set of parentheses separately. - What happens to a difference when both numbers increase by the same amount? - What happens to a sum when both addends increase by \(15\)?

Solution

1. In \(560 - 140\), increasing both numbers by \(15\) does not change the difference. 2. In \(320 + 180\), increasing both addends by \(15\) increases the sum by \(15 + 15 = 30\). 3. Therefore, the value of the entire expression increases by \(30\). The original value is \(920\), and the new value is \(950\).

Answer

The value increases by \(30\).
5186387
Consider the expression \((1200 + 800) - (600 - 200)\). Determine how its value changes if each of the four numbers is increased by \(25\). Explain your reasoning from the structure of the expression.

Hints

- Determine how the first set of parentheses changes. - Determine how the second set of parentheses changes. - Use those two changes to determine the change in the entire expression.

Solution

1. Increasing both addends in \(1200 + 800\) by \(25\) increases that sum by \(50\). 2. Increasing both numbers in \(600 - 200\) by \(25\) leaves that difference unchanged. 3. Therefore, the value of the entire expression increases by \(50\). The original value is \(1600\), and the new value is \(1650\).

Answer

The value increases by \(50\).
5187167
At an Arctic weather station, the outdoor temperature is \(-46\,^\circ\text{C}\) early in the morning. By noon, it has risen to \(-19\,^\circ\text{C}\). By how many degrees Celsius did the temperature increase?

Hints

- Picture the temperatures on a vertical number line. - Did the temperature increase or decrease? - How can you find the distance between the two values?

Solution

1. Subtract the initial temperature from the final temperature: \(-19-(-46)\). 2. Calculate: \(-19+46=27\).

Answer

The temperature increased by \(27\,^\circ\text{C}\).
5187177
An unmanned research submarine is at \(-780\,\text{m}\) relative to sea level. It must rise to a maintenance station at \(-125\,\text{m}\). How many meters must the submarine rise?

Hints

- What does \(0\) represent in this situation? - Sketch the two positions on a vertical number line. - Which operation finds the change from the starting position to the ending position?

Solution

1. Find the change from the starting position to the ending position: \(-125-(-780)\). 2. Calculate: \(-125+780=655\).

Answer

The submarine must rise \(655\,\text{m}\).
5187217
Solve for \(x\). a) \(x+85=42\) b) \(-14+x=36\) c) \(x-55=-19\) d) \(-72+x=-110\)

Hints

- Undo the operation applied to \(x\). - Subtract a known addend to find the missing one. - Check each solution by substitution.

Solution

1. In part a), \(x=42-85=-43\). 2. In part b), \(x=36-(-14)=50\). 3. In part c), \(x=-19+55=36\). 4. In part d), \(x=-110-(-72)=-38\).

Answer

a) \(x=-43\) b) \(x=50\) c) \(x=36\) d) \(x=-38\)
5188957
Evaluate each expression. a) \(72-115+43\) b) \(-56+88-32\) c) \(105-(-45)-200\)

Hints

- Work from left to right. - Distinguish signs attached to numbers from operation signs. - Subtracting a negative number is the same as adding a positive number.

Solution

1. In a), \(72-115=-43\), and \(-43+43=0\). 2. In b), \(-56+88=32\), and \(32-32=0\). 3. In c), \(105-(-45)-200=105+45-200=-50\).

Answer

a) \(0\) b) \(0\) c) \(-50\)
5190107
Evaluate \(-135-(62-140)\).

Hints

- Evaluate the parentheses first. - Subtracting a negative number becomes addition. - Compare the absolute values to determine the sign of the final sum.

Solution

1. Evaluate inside the parentheses: \(62-140=-78\). 2. Substitute the result: \(-135-(-78)=-135+78\). 3. Evaluate: \(-135+78=-57\).

Answer

\(-57\)
5193017
Evaluate \(-135+48+35-18\) by regrouping efficiently.

Hints

- Look for pairs that make convenient tens or hundreds. - Keep each sign attached to its number when reordering. - Add the partial sums last.

Solution

1. Regroup as \((-135+35)+(48-18)\). 2. The partial sums are \(-100\) and \(30\). 3. Therefore, the value is \(-100+30=-70\).

Answer

\(-70\)
5193027
Simplify the operation signs and signs attached to numbers, and then evaluate: \(-12-(-18)+(-25)\).

Hints

- Two consecutive negative signs become addition. - Adding a negative number is subtraction. - Rewrite the expression before calculating.

Solution

1. Rewrite the expression as \(-12+18-25\). 2. Evaluate: \(-12+18=6\), and \(6-25=-19\).

Answer

\(-19\)
5199557
Imagine moving on a number line. Subtracting a positive number means moving left. Write and evaluate the subtraction for each situation. a) Start at \(20\) and subtract \(55\). b) Start at \(-15\) and subtract \(30\).

Hints

- Subtracting a positive number moves left on a number line. - You can first move to zero and then count the remaining distance. - Starting negative and subtracting a positive number makes the result more negative.

Solution

1. In a), move \(55\) units left from \(20\): \(20-55=-35\). 2. In b), move \(30\) units left from \(-15\): \(-15-30=-45\).

Answer

a) \(20-55=-35\) b) \(-15-30=-45\)
5217577
Evaluate each expression. a) \((-75)+(-25)\) b) \((-75)-(-25)\) c) \(75+(-25)\) d) \(75-(-25)\)

Hints

- Adding a negative number is the same as subtraction. - Subtracting a negative number is the same as addition. - Use a number line to check whether each value moves left or right.

Solution

1. In a), \(-75+(-25)=-100\). 2. In b), \(-75-(-25)=-75+25=-50\). 3. In c), \(75+(-25)=50\). 4. In d), \(75-(-25)=75+25=100\).

Answer

a) \(-100\) b) \(-50\) c) \(50\) d) \(100\)
5217677
Find the missing value in each addition equation. a) \(-18+30=\square\) b) \(\square+(-12)=-20\) c) \(-45+\square=-15\)

Hints

- Write an addition equation for each part. - Use subtraction to find a missing addend. - Check each completed equation.

Solution

1. In a), \(-18+30=12\). 2. In b), \(A+(-12)=-20\), so \(A=-20-(-12)=-8\). 3. In c), \(-45+B=-15\), so \(B=-15-(-45)=30\).

Answer

a) \(12\) b) \(-8\) c) \(30\)
5217697
Rewrite subtraction as addition of the opposite. Then use the commutative and associative properties to show a grouping that creates \(100\), and evaluate. \(47 - 19 + 53\)

Hints

- First express every operation as addition of a signed number. - Which two positive addends form a convenient benchmark number? - Show the property-based rewrite in the answer.

Solution

1. Rewrite: \(47+(-19)+53\). 2. Reorder and regroup: \((47+53)+(-19)\). 3. Evaluate: \(100-19=81\).

Answer

\(47+(-19)+53=(47+53)+(-19)=100-19=81\)
5225527
Use sea level as \(0\,\text{m}\). a) A submarine is \(250\,\text{m}\) below sea level. Write its position as an integer. b) A climber is on a peak \(1850\,\text{m}\) above sea level. Write the elevation as an integer. c) A diver is at \(-15\,\text{m}\) and descends another \(10\,\text{m}\). Write the new position as an integer. d) Find the vertical distance between a bird at \(+12\,\text{m}\) and a fish at \(-8\,\text{m}\).

Hints

- Positions below the reference level are negative. - Picture a vertical number line with sea level at zero. - If you are already below zero and move lower, does the number become greater or less? - To find a distance that crosses zero, combine the distances from each point to zero.

Solution

1. A position below sea level is negative, so the submarine's position is \(-250\). 2. A position above sea level is positive, so the climber's elevation is \(+1850\). 3. Moving \(10\,\text{m}\) lower from \(-15\,\text{m}\) gives \(-25\,\text{m}\). 4. The distance from \(+12\) to \(-8\) is \(12+8=20\,\text{m}\).

Answer

a) \(-250\) b) \(+1850\) c) \(-25\) d) \(20\,\text{m}\)
5225687
A parking garage labels street level as Level \(0\). Levels below street level are negative, and levels above street level are positive. a) A car is parked on Level \(-3\). Describe its location. b) A visitor parks on Level \(-1\) and takes the elevator up \(4\) levels. On which level does the visitor exit? c) Later, the visitor rides directly from that level to Level \(-2\). Represent the elevator's change in level with a signed number.

Hints

- Picture the levels on a vertical number line. - Moving up is a positive change, and moving down is a negative change. - Count the change from Level \(3\) through Level \(0\) to Level \(-2\).

Solution

1. Level \(-3\) is \(3\) levels below street level. 2. Moving up \(4\) levels gives \(-1+4=3\), so the visitor exits on Level \(3\). 3. The change from Level \(3\) to Level \(-2\) is \(-2-3=-5\), representing a trip down \(5\) levels.

Answer

a) The car is \(3\) levels below street level. b) The visitor exits on Level \(3\). c) The change is \(-5\), meaning \(5\) levels down.
5225727
A mountain weather station tracks temperature changes during the day. In the morning, the temperature rises by \(x\,^{\circ}\text{C}\). In the afternoon, it falls by \(y\,^{\circ}\text{C}\). a) Write an expression for the net temperature change. b) Find the net change on each day. 1) Day 1: \(x=5.4\), \(y=3.1\) 2) Day 2: \(x=2.8\), \(y=4.5\) c) Explain what the sign of each result means.

Hints

- Think about the direction of each change on a thermometer. - Subtract the amount of the decrease from the amount of the increase. - Interpret a negative result as an overall drop.

Solution

1. The net change is the increase minus the decrease: \(x-y\). 2. Day 1: \(5.4-3.1=2.3\), so the temperature rises by \(2.3\,^{\circ}\text{C}\) overall. 3. Day 2: \(2.8-4.5=-1.7\), so the temperature falls by \(1.7\,^{\circ}\text{C}\) overall. 4. A positive result indicates a net warming, and a negative result indicates a net cooling.

Answer

a) \(x-y\) b) Day 1: \(2.3\,^{\circ}\text{C}\); Day 2: \(-1.7\,^{\circ}\text{C}\) c) A positive value means the day ends warmer than it began; a negative value means it ends cooler.
5225757
A class fund starts the month with some savings. During the month, it earns \(\$e\) from a bake sale and spends \(\$a\) on art supplies. a) Write an expression for the change in the fund balance. b) Find the change when \(e=54\) and \(a=38\). Explain the result. c) Find the change when \(e=25\) and \(a=42\). Compare the ending balance with the starting balance.

Hints

- Income adds to the fund, while spending removes money. - If spending is greater than income, the change is negative. - A positive value represents an increase; a negative value represents a decrease.

Solution

1. The change is income minus spending: \(e-a\). 2. For \(e=54\) and \(a=38\), \(54-38=16\). The balance increases by \(\$16\). 3. For \(e=25\) and \(a=42\), \(25-42=-17\). The balance ends \(\$17\) lower than it began.

Answer

a) \(e-a\) b) \(\$16\); the balance increases by \(\$16\). c) \(-\$17\); the ending balance is \(\$17\) less than the starting balance.
5225767
A reservoir helps regulate a river. Each hour, \(z\,\text{m}^3\) of water flows into the reservoir, and \(a\,\text{m}^3\) flows out through the gates. a) Write an expression for the change in water volume after one hour. b) What does a value of \(0\) mean in this context? c) What does a negative value mean? d) Find the hourly change when \(z=1250\) and \(a=1310\).

Hints

- Think of a container with both an inlet and an outlet. - The level stays constant when inflow equals outflow. - The volume decreases when more water leaves than enters.

Solution

1. The net change is inflow minus outflow: \(z-a\). 2. If \(z-a=0\), inflow equals outflow, so the water volume stays constant. 3. If \(z-a<0\), outflow exceeds inflow, so the water volume decreases. 4. For \(z=1250\) and \(a=1310\), \(1250-1310=-60\). The volume decreases by \(60\,\text{m}^3\) per hour.

Answer

a) \(z-a\) b) The water volume remains constant. c) The reservoir's water volume is decreasing. d) \(-60\,\text{m}^3\) per hour
5225797
Tim is saving for a skateboard that costs \(\$85\). He has already saved \(\$y\). a) Write an expression for the signed difference “price minus savings.” b) Evaluate the expression for \(y=62\) and for \(y=90\). Explain what each result means.

Hints

- Consider what happens when the savings exceed the price. - Start by imagining a specific savings amount below \(85\). - Interpret the sign of the result in the context.

Solution

1. The signed difference is \(85-y\). 2. For \(y=62\), \(85-62=23\). Tim still needs \(\$23\). 3. For \(y=90\), \(85-90=-5\). Tim has saved \(\$5\) more than the price.

Answer

a) \(85-y\) b) For \(y=62\): \(23\), so Tim needs \(\$23\) more. For \(y=90\): \(-5\), so Tim has \(\$5\) extra.
5225807
A hiker is at an elevation of \(h\) feet. The destination lodge is at an elevation of \(4000\) feet. a) Write an expression for the signed elevation difference “destination elevation minus current elevation.” b) Evaluate the difference for \(h=3200\) and for \(h=4500\). Interpret each result as an uphill or downhill change.

Hints

- Picture elevation on a vertical number line. - A destination above the current elevation gives a positive difference. - A negative result indicates that the destination is below the current elevation.

Solution

1. The signed elevation difference is \(4000-h\). 2. For \(h=3200\), \(4000-3200=800\). The hiker must climb \(800\) feet. 3. For \(h=4500\), \(4000-4500=-500\). The hiker is \(500\) feet above the lodge and must descend \(500\) feet.

Answer

a) \(4000-h\) b) For \(h=3200\): \(800\) feet uphill. For \(h=4500\): \(-500\) feet, meaning \(500\) feet downhill.
5226017
A bank account has a current balance of \(\$x\). A bill of \(\$y\) is withdrawn. a) Write an expression for the new account balance. b) Find the balance when \(x=125\) and \(y=40\). c) Find the balance when \(x=50\) and \(y=85\). Explain what a negative result means for the account holder.

Hints

- A withdrawal decreases the account balance. - Spending more than the current balance produces a negative result.

Solution

1. Subtract the withdrawal from the current balance: \(x-y\). 2. For \(x=125\) and \(y=40\), \(125-40=85\). 3. For \(x=50\) and \(y=85\), \(50-85=-35\). 4. A negative balance means the account is overdrawn by that amount.

Answer

a) \(x-y\) b) \(\$85\) c) \(-\$35\); the account is overdrawn by \(\$35\).
5226247
Evaluate by rearranging and grouping the addends. Show the reordered expression, the two grouped partial sums, and the final value. \(17.4 + (-5.9) + 2.6 + (-4.1)\) Which property allows you to change the order of the addends?

Hints

- Look for decimal pairs whose sum is a whole number. - Separate changing the order from changing the grouping. - The answer must show both the reordered expression and the resulting groups.

Solution

1. Reorder: \(17.4+2.6+(-5.9)+(-4.1)\). 2. Group: \((17.4+2.6)+((-5.9)+(-4.1))\). 3. The partial sums are \(20\) and \(-10\), so the final value is \(10\). 4. The commutative property of addition allows the order to change.

Answer

Reordered expression: \(17.4+2.6+(-5.9)+(-4.1)\) Partial sums: \(20\) and \(-10\) Final value: \(10\) Property: commutative property of addition
5226277
Find each sum of rational numbers. 1) \((-17)+(+25)+(-13)\) 2) \((+4.8)+(-6.3)+(-2.5)\) 3) \(\left(-\frac{3}{10}\right)+0.75\) 4) \(\left(-2 \frac{1}{4}\right)+\left(-1 \frac{5}{8}\right)\)

Hints

- Pay close attention to signs when adding rational numbers. - You can combine several addends step by step. - When fractions and decimals appear together, convert them to a common form if that makes the calculation easier. - Fractions need a common denominator before they can be added.

Solution

1. For 1), \(-17+25=8\), then \(8+(-13)=-5\). 2. For 2), \(4.8+(-6.3)=-1.5\), then \(-1.5+(-2.5)=-4\). 3. For 3), \(-\frac{3}{10}=-0.3\), so \(-0.3+0.75=0.45\). Equivalently, \(-\frac{6}{20}+\frac{15}{20}=\frac{9}{20}\). 4. For 4), write the mixed numbers with denominator \(8\): \(-2 \frac{2}{8}+\left(-1 \frac{5}{8}\right)=-3 \frac{7}{8}\).

Answer

1) \(-5\) 2) \(-4\) 3) \(0.45\), or \(\frac{9}{20}\) 4) \(-3 \frac{7}{8}\)
5226407
For each pair of values, find \(c\) so that \(a+b+c=0\). a) \(a=-8.5\), \(b=3\frac{1}{4}\) b) \(a=\frac{2}{3}\), \(b=-\frac{1}{6}\) c) \(a=15.7\), \(b=-20\)

Hints

- First find \(a+b\). - The total is zero when \(c\) is the opposite of \(a+b\). - Use a common denominator when adding or subtracting fractions.

Solution

1. For a), \(a+b=-8.5+3.25=-5.25\). The opposite of \(-5.25\) is \(5.25\), so \(c=5.25\). 2. For b), \(a+b=\frac{2}{3}-\frac{1}{6}=\frac{4}{6}-\frac{1}{6}=\frac{1}{2}\). Its opposite is \(-\frac{1}{2}\), so \(c=-\frac{1}{2}\). 3. For c), \(a+b=15.7-20=-4.3\). Its opposite is \(4.3\), so \(c=4.3\).

Answer

a) \(c=5.25\) b) \(c=-\frac{1}{2}\) c) \(c=4.3\)
5226417
A weather station records morning and noon temperatures. Calculate each temperature change as \(\text{noon temperature}-\text{morning temperature}\), and enter a positive number, a negative number, or zero in the table. <table> <thead> <tr> <th>Day</th> <th>Morning temperature</th> <th>Noon temperature</th> <th>Temperature change</th> </tr> </thead> <tbody> <tr> <td>Monday</td> <td>\(+2\,^\circ\text{C}\)</td> <td>\(-3\,^\circ\text{C}\)</td> <td></td> </tr> <tr> <td>Tuesday</td> <td>\(-5\,^\circ\text{C}\)</td> <td>\(+1\,^\circ\text{C}\)</td> <td></td> </tr> <tr> <td>Wednesday</td> <td>\(-4\,^\circ\text{C}\)</td> <td>\(-4\,^\circ\text{C}\)</td> <td></td> </tr> <tr> <td>Thursday</td> <td>\(0\,^\circ\text{C}\)</td> <td>\(-6\,^\circ\text{C}\)</td> <td></td> </tr> </tbody> </table>

Hints

- Decide whether each temperature became warmer or colder. - A decrease is represented by a negative change. - An increase is represented by a positive change. - No change is represented by zero.

Solution

1. Monday: \(-3-2=-5\,^\circ\text{C}\). 2. Tuesday: \(1-(-5)=+6\,^\circ\text{C}\). 3. Wednesday: \(-4-(-4)=0\,^\circ\text{C}\). 4. Thursday: \(-6-0=-6\,^\circ\text{C}\).

Answer

Monday: \(-5\,^\circ\text{C}\) Tuesday: \(+6\,^\circ\text{C}\) Wednesday: \(0\,^\circ\text{C}\) Thursday: \(-6\,^\circ\text{C}\)
5226617
Evaluate each expression. 1) \((-18)-(+12)-(-20)\) 2) \((-5.4)-(-2.6)-(+1.2)\) 3) \((+3.7)-(+8.7)-(-5)\)

Hints

- What happens to the sign of a number in parentheses when a minus sign is directly before it? - Can you rewrite subtraction as addition of the opposite? - After rewriting, work from left to right. - Keep decimal place values aligned and track the signs at each step.

Solution

1. For 1), rewrite subtraction of a negative as addition: \(-18-12+20=-30+20=-10\). 2. For 2), \(-5.4-(-2.6)-1.2=-5.4+2.6-1.2=-2.8-1.2=-4\). 3. For 3), \(3.7-8.7-(-5)=3.7-8.7+5=-5+5=0\).

Answer

1) \(-10\) 2) \(-4\) 3) \(0\)
5228907
For each sum, write the grouping that creates an additive inverse or a whole-number partial sum, then evaluate. 1) \((-34) + 57 + 34\) 2) \(12.4 + (-6.8) + (-3.2)\) 3) \(-2\frac{1}{5} + 8 - 1\frac{4}{5}\)

Hints

- Search for opposites or terms that combine to an integer. - Keep each sign with its term when grouping. - Include the grouping that makes the arithmetic short.

Solution

1. \((-34+34)+57=57\). 2. \(12.4+(-6.8-3.2)=12.4-10=2.4\). 3. \(\left(-2\frac{1}{5}-1\frac{4}{5}\right)+8=-4+8=4\).

Answer

1) \((-34+34)+57=57\) 2) \(12.4+(-6.8-3.2)=2.4\) 3) \(\left(-2\frac{1}{5}-1\frac{4}{5}\right)+8=4\)
5279437
Calculate \(x-y\) for each pair. a) \(x=-2.4\), \(y=5.6\) b) \(x=\frac{3}{5}\), \(y=-\frac{7}{10}\) c) \(x=-1\frac{1}{2}\), \(y=-2\frac{1}{4}\) d) \(x=-0.75\), \(y=0.75\)

Hints

- Subtracting a negative number is the same as adding its opposite. - Use a common denominator before subtracting fractions. - For decimal differences, think about the expected sign before calculating.

Solution

1. For a), \(x-y=-2.4-5.6=-8.0\). 2. For b), \(x-y=\frac{3}{5}-\left(-\frac{7}{10}\right)=\frac{6}{10}+\frac{7}{10}=\frac{13}{10}=1\frac{3}{10}\). 3. For c), \(x-y=-1\frac{1}{2}-\left(-2\frac{1}{4}\right)=-\frac{6}{4}+\frac{9}{4}=\frac{3}{4}\). 4. For d), \(x-y=-0.75-0.75=-1.5\).

Answer

a) \(-8.0\) b) \(1\frac{3}{10}\), or \(1.3\) c) \(\frac{3}{4}\), or \(0.75\) d) \(-1.5\)
5319137
The number lines show integer jumps. Find each number marked with a red question mark.
Figure for problem 531913

Hints

- A positive jump goes right, and a negative jump goes left. - To find a missing start, undo the jump from the ending value. - For multiple jumps, find each landing in order.

Solution

1. In a), \(-24+35=11\). 2. In b), the starting value satisfies \(x+(-42)=-15\), so \(x=-15-(-42)=27\). 3. In c), \(15+(-23)=-8\), and then \(-8+12=4\).

Answer

a) \(11\) b) \(27\) c) First value: \(-8\); second value: \(4\)
5319157
The number lines show integer jumps. Find the number marked with a red question mark in each part. a) Find the ending value. b) Find the starting value. c) Find the missing intermediate value.
Figure for problem 531915

Hints

- Positive jumps move right. - To find a missing start, subtract the jump from the ending value. - For multiple jumps, calculate the first landing before the next jump.

Solution

1. In a), \(-14+32=18\). 2. In b), the starting value satisfies \(x+45=15\), so \(x=15-45=-30\). 3. In c), \(-28+18=-10\). The check \(-10+22=12\) matches the shown ending value.

Answer

a) \(18\) b) \(-30\) c) \(-10\)
5319377
Find the number marked with a red question mark on each number line.
Figure for problem 531937

Hints

- Follow the direction and signed size of each jump. - Add the jump when the ending value is missing. - Subtract the jump when the starting value is missing.

Solution

1. In a), \(-15+(-18)=-33\). 2. In b), \(x+(-24)=-32\), so \(x=-32-(-24)=-8\). 3. In c), \(-45+60=15\). 4. In d), \(x+(-35)=-23\), so \(x=-23-(-35)=12\).

Answer

a) \(-33\) b) \(-8\) c) \(15\) d) \(12\)
5350457
A school snack stand tracks its monthly finances. The bar chart shows the profit (positive values) or loss (negative values), in dollars, for the first six months of the year. a) Find the total profit or loss after the six months. b) Were there more months with a profit or more months with a loss?
Figure for problem 535045

Hints

- Treat profits as positive values and losses as negative values. - Use the zero line to tell whether each month shows a profit or a loss. - Count the bars above zero and compare that count with the bars below zero.

Solution

1. Read the monthly values from the chart: Jan. \(\$35\), Feb. \(-\$20\), Mar. \(\$55\), Apr. \(-\$45\), May \(\$90\), and Jun. \(\$35\). 2. For a), add the signed values: \(35-20+55-45+90+35=150\). The six-month total is a profit of \(\$150\). 3. For b), four months have positive values and two months have negative values, so there were more months with a profit.

Answer

a) A profit of \(\$150\) b) More months had a profit: \(4\) profit months and \(2\) loss months.
5351537
Find each ending value on the number lines. Write an addition equation for each jump.
Figure for problem 535153

Hints

- Identify the starting value. - A rightward jump is positive, and a leftward jump is negative. - Add the signed jump to the start.

Solution

1. In a), \(15+(-22)=-7\). 2. In b), \(-12+18=6\). 3. In c), \(-10+(-15)=-25\).

Answer

a) \(15+(-22)=-7\) b) \(-12+18=6\) c) \(-10+(-15)=-25\)
5351697
Find each missing value on the number lines.
Figure for problem 535169

Hints

- Add the jump when the ending value is unknown. - Use the inverse operation when the starting value is unknown. - Check the direction of each signed jump.

Solution

1. In a), \(-125+75=-50\). 2. In b), \(240+(-380)=-140\). 3. In c), \(x+(-150)=-400\), so \(x=-250\). 4. In d), \(x+220=50\), so \(x=-170\).

Answer

a) \(-50\) b) \(-140\) c) \(-250\) d) \(-170\)
5351877
Find the missing ending value on each number line.
Figure for problem 535187

Hints

- Read the starting number and signed jump. - A positive jump moves right. - Add the jump to find the ending value.

Solution

1. In a), \(-15+24=9\). 2. In b), \(-8+15=7\). 3. In c), \(-30+55=25\).

Answer

a) \(9\) b) \(7\) c) \(25\)
5351897
Find the missing ending value for each negative jump.
Figure for problem 535189

Hints

- Read the signed size printed on each jump. - A negative jump moves left from the displayed starting value. - Combine the starting value with the signed jump to find the landing.

Solution

1. a) \(20+(-45)=-25\). 2. b) \(12+(-30)=-18\). 3. c) \(5+(-15)=-10\).

Answer

a) \(-25\) b) \(-18\) c) \(-10\)
5352547
Use the number line to find the missing ending value.
Figure for problem 535254

Hints

- Read the starting value and signed jump from the diagram. - Combine the signed jump with the start to locate the landing.

Solution

1. The diagram starts at \(-42\) and shows a jump of \(+65\). 2. \(-42+65=23\).

Answer

\(23\)
5352577
The number line shows the peak elevation and the vertical distance from the base to the peak of an underwater mountain. How deep is the base of the mountain below sea level?
Figure for problem 535257

Hints

- Read the peak elevation and the mountain's vertical height from the number line. - The base is lower than the peak, so represent the change with subtraction. - Report depth as a positive distance below sea level.

Solution

1. The number line shows the peak at \(-50\,\text{m}\). 2. The base is \(2600\,\text{m}\) lower, so its elevation is \(-50-2600=-2650\,\text{m}\). 3. Therefore, the base is \(2650\,\text{m}\) below sea level.

Answer

The base is \(2650\,\text{m}\) below sea level.
5352587
The number-line diagram shows Max’s starting bank account balance and a deposit. What is his new balance?
Figure for problem 535258

Hints

- Read the starting balance from the diagram. - A deposit is represented by a positive jump. - Combine the starting balance with the change shown by the jump.

Solution

1. The diagram shows a starting balance of \(-\$15\) and a positive change of \(\$40\). 2. Add the deposit: \(-15+40=25\).

Answer

His new balance is \(\$25\).
5545517
The number line shows the change produced by subtracting \(-5\) from the displayed starting value. a) Find the missing ending value. b) Write the subtraction equation represented by the diagram. c) Explain why the move is to the right even though the operation is subtraction.
Figure for problem 554551

Hints

- Read the start and signed jump from the diagram. - Compare the diagram's positive jump with the effect of subtracting a negative number. - Your explanation should connect the subtraction expression to the direction shown.

Solution

1. The displayed start is \(-3\), and the diagram shows a jump of \(+5\). 2. Subtracting \(-5\) is equivalent to adding \(+5\), so \(-3-(-5)=-3+5=2\). 3. Adding a positive value moves right on the number line.

Answer

a) \(2\) b) \(-3-(-5)=2\) c) Subtracting \(-5\) is equivalent to adding \(5\), so the movement is to the right.
5545527
Point \(A\) is a starting position and point \(B\) is an ending position on the number line. a) Read \(A\) and \(B\) as fractions in simplest form. b) Write the signed change from \(A\) to \(B\) as \(B-A\), and evaluate it. c) Does the sign of the change match the direction of movement? Explain.
Figure for problem 554552

Hints

- Determine the value of one small tick interval from the labeled integers. - Read both coordinates before forming the signed difference. - Relate the sign of \(B-A\) to the direction from the starting point to the ending point.

Solution

1. The ticks divide each unit into eighths, so \(A=-\frac{5}{8}\) and \(B=\frac{1}{4}\). 2. \(B-A=\frac{1}{4}-\left(-\frac{5}{8}\right)=\frac{2}{8}+\frac{5}{8}=\frac{7}{8}\). 3. The change is positive, matching movement to the right from \(A\) to \(B\).

Answer

a) \(A=-\frac{5}{8}\), \(B=\frac{1}{4}\) b) \(B-A=\frac{7}{8}\) c) Yes. The positive change matches movement to the right.
5103177
Find the fraction exactly halfway between \(\frac{2}{3}\) and \(\frac{4}{5}\). Write your answer in simplest form.

Hints

- The number halfway between two values is their mean. - Use a common denominator before adding the fractions. - Divide the sum by \(2\).

Solution

1. The midpoint is the mean: \(\frac{\frac{2}{3}+\frac{4}{5}}{2}\). 2. Use a common denominator: \(\frac{2}{3}=\frac{10}{15}\) and \(\frac{4}{5}=\frac{12}{15}\). 3. Add: \(\frac{10}{15}+\frac{12}{15}=\frac{22}{15}\). 4. Divide by \(2\): \(\frac{22}{15}\div2=\frac{22}{15}\cdot\frac{1}{2}=\frac{11}{15}\).

Answer

\(\frac{11}{15}\)
5103607
Compare the structure of expressions \(A\) and \(B\) before calculating. \(A=-15+(30-45)\) \(B=-15+30-45\) 1) Find the value of each expression. 2) What rule about removing parentheses is illustrated here? 3) How would \(B\) need to change if \(A\) were \(-15-(30-45)\) instead?

Hints

- Compare the signs inside the parentheses with the signs after the parentheses are removed. - Think about the difference between a plus sign and a minus sign before parentheses. - State the pattern in your own words.

Solution

1. \(A=-15+(-15)=-30\), and \(B=-15+30-45=-30\). 2. A plus sign before parentheses allows the parentheses to be removed without changing the signs inside. 3. With a minus sign before the parentheses, the signs inside change when the parentheses are removed: \(-15-30+45\).

Answer

1) \(A=-30\), \(B=-30\) 2) A plus sign before parentheses allows the parentheses to be removed without changing the signs inside. 3) \(-15-30+45\)
5103697
Point \(P\) is at \(-3.4\) on a number line, and point \(Q\) is at \(1.2\). a) Find the number \(m\) exactly halfway between \(P\) and \(Q\). b) Is \(m\) an integer? Is it a rational number? c) List all integers strictly between \(-3.4\) and \(1.2\).

Hints

- Find the midpoint by adding the endpoints and dividing by \(2\). - A terminating decimal can be written as a fraction. - Move from left to right on the number line and record each integer between the endpoints.

Solution

1. Find the midpoint: \(m=\frac{-3.4+1.2}{2}=\frac{-2.2}{2}=-1.1\). 2. The value \(-1.1\) is not an integer. It is rational because \(-1.1=-\frac{11}{10}\). 3. The integers satisfying \(-3.4<z<1.2\) are \(-3, -2, -1, 0, 1\).

Answer

a) \(m=-1.1\) b) \(m\) is not an integer, but it is rational. c) \(-3, -2, -1, 0, 1\)
5104267
Compare \(\frac{19}{8}\) and \(2.37\). a) Which number is greater? b) Find the decimal exactly halfway between the two numbers.

Hints

- Convert the fraction to a decimal. - Add a trailing zero when comparing \(2.37\) with a number having three decimal places. - The number halfway between two values is their average.

Solution

1. Convert the fraction: \(\frac{19}{8}=2.375\). 2. Compare \(2.375\) and \(2.370\). Therefore, \(\frac{19}{8}>2.37\). 3. Find the midpoint: \(\frac{2.375+2.37}{2}=\frac{4.745}{2}=2.3725\).

Answer

a) \(\frac{19}{8}\) is greater. b) \(2.3725\)
5104477
Find the number exactly halfway between each pair. a) Between \(\frac{5}{8}\) and \(0.63\). Give the answer as a decimal. b) Between \(0.7\) and \(0.71\). Give the answer as a fraction in simplest form.

Hints

- The number exactly halfway between two values is their average. - Convert the fraction in part a) to a decimal. - In part b), use equivalent fractions with a denominator large enough to leave a numerator between the two endpoints.

Solution

1. For a), \(\frac{5}{8}=0.625\). The midpoint is \(\frac{0.625+0.63}{2}=\frac{1.255}{2}=0.6275\). 2. For b), write the endpoints as fractions: \(0.7=\frac{140}{200}\) and \(0.71=\frac{142}{200}\). 3. The fraction halfway between them is \(\frac{141}{200}\), which is already in simplest form.

Answer

a) \(0.6275\) b) \(\frac{141}{200}\)
5104577
Find the decimal exactly halfway between \(\frac{1}{5}\) and \(\frac{1}{4}\). Then give one decimal greater than \(\frac{1}{5}\) but less than your midpoint.

Hints

- Convert both fractions to decimals. - Find the average of the two endpoint values. - Choose a value strictly between the lower endpoint and the midpoint.

Solution

1. Convert the fractions: \(\frac{1}{5} = 0.2\) and \(\frac{1}{4} = 0.25\). 2. Average the endpoints: \(\frac{0.2 + 0.25}{2} = \frac{0.45}{2} = 0.225\). 3. Any decimal strictly between \(0.2\) and \(0.225\) works, such as \(0.21\).

Answer

The midpoint is \(0.225\). One possible additional decimal is \(0.21\).
5104587
Consider \(\frac{3}{8}\), \(0.3\), \(\frac{2}{5}\), and \(0.38\). a) Convert the fractions to decimals and order all four numbers from least to greatest. b) Give one decimal strictly between the two greatest values from part a).

Hints

- Convert both fractions to decimals first. - Compare tenths, hundredths, and thousandths in order. - Write \(0.4\) as \(0.40\) if that helps you find a number between the last two values.

Solution

1. Convert the fractions: \(\frac{3}{8} = 0.375\) and \(\frac{2}{5} = 0.4\). 2. Compare the decimals: \(0.3 < 0.375 < 0.38 < 0.4\). 3. The two greatest values are \(0.38\) and \(0.4\). One decimal between them is \(0.39\).

Answer

a) \(0.3 < \frac{3}{8} < 0.38 < \frac{2}{5}\) b) Answers will vary. One possible answer is \(0.39\).
5105647
Calculate the expression and write the result as a fraction in simplest form: \(0.4 + \frac{1}{3} - 15\%\) What problem occurs if you try to calculate this expression exactly using only terminating decimals and no fractions?

Hints

- Try rewriting the decimal and the percent as fractions first. - Find a common denominator for the fractions in the expression. - Think about whether every rational number has a terminating decimal representation.

Solution

1. Rewrite the decimal and percent as fractions: \(0.4 = \frac{2}{5}\) and \(15\% = \frac{3}{20}\). 2. The expression becomes \(\frac{2}{5} + \frac{1}{3} - \frac{3}{20}\). 3. Use a common denominator of \(60\): \(\frac{24}{60} + \frac{20}{60} - \frac{9}{60} = \frac{35}{60} = \frac{7}{12}\). 4. The fraction \(\frac{1}{3}\) has the repeating decimal representation \(0.\overline{3}\), so it cannot be represented exactly by a terminating decimal. Any terminating decimal replacement would be an approximation.

Answer

The result is \(\frac{7}{12}\). Using only terminating decimals cannot give an exact calculation because \(\frac{1}{3}\) has a repeating decimal representation.
5106037
Find the rational number \(x\) exactly halfway between \(-\frac{1}{2}\) and \(0.3\). Then determine whether \(x\) is greater than or less than \(-\frac{1}{8}\). Justify your comparison.

Hints

- The number halfway between two values is their average. - Convert the fractions and decimals to a common form. - On a number line, the value farther right is greater.

Solution

1. Convert \(-\frac{1}{2}\) to \(-0.5\). 2. Find the midpoint: \(x=\frac{-0.5+0.3}{2}=\frac{-0.2}{2}=-0.1\). 3. Convert the comparison value: \(-\frac{1}{8}=-0.125\). 4. Since \(-0.1>-0.125\), \(x> -\frac{1}{8}\).

Answer

\(x=-0.1\), and \(x> -\frac{1}{8}\).
5106117
Calculate each expression. Write each answer as a fraction in simplest form or as a mixed number. a) \(1 \frac{3}{8} + \frac{7}{8} - \frac{5}{8}\) b) \(\frac{4}{15} - \left(\frac{7}{15} + \frac{12}{15}\right)\) c) \(2 \frac{1}{12} - \frac{5}{12} - \frac{11}{12}\)

Hints

- How can you rewrite a mixed number before calculating? - Remember to evaluate the expression inside parentheses first. - A subtraction involving fractions can have a negative result. - At the end, simplify the fraction or rewrite an improper fraction as a mixed number when appropriate.

Solution

1. For a), rewrite \(1 \frac{3}{8}\) as \(\frac{11}{8}\). Then \(\frac{11+7-5}{8} = \frac{13}{8} = 1 \frac{5}{8}\). 2. For b), evaluate the parentheses first: \(\frac{7}{15} + \frac{12}{15} = \frac{19}{15}\). Then \(\frac{4}{15} - \frac{19}{15} = -\frac{15}{15} = -1\). 3. For c), rewrite \(2 \frac{1}{12}\) as \(\frac{25}{12}\). Then \(\frac{25-5-11}{12} = \frac{9}{12} = \frac{3}{4}\).

Answer

a) \(1 \frac{5}{8}\) b) \(-1\) c) \(\frac{3}{4}\)
5106217
Calculate the value of the nested expression: \(\frac{2}{5} - \left(\frac{1}{2} - \left(\frac{3}{10} + \frac{1}{4}\right)\right)\)

Hints

- With nested parentheses, work from the innermost parentheses outward. - Subtracting a negative number is the same as adding its opposite. - Find a useful common denominator at each step.

Solution

1. Start with the innermost parentheses: \(\frac{3}{10} + \frac{1}{4} = \frac{6}{20} + \frac{5}{20} = \frac{11}{20}\). 2. Evaluate the next parentheses: \(\frac{1}{2} - \frac{11}{20} = \frac{10}{20} - \frac{11}{20} = -\frac{1}{20}\). 3. Subtract the negative fraction: \(\frac{2}{5} - \left(-\frac{1}{20}\right) = \frac{2}{5} + \frac{1}{20}\). 4. Use denominator \(20\): \(\frac{8}{20} + \frac{1}{20} = \frac{9}{20}\).

Answer

\(\frac{9}{20}\)
5106297
Compare the results of calculations \(A\) and \(B\). Insert \(<\), \(>\), or \(=\), and justify your answer with calculations. \(A=5 \frac{1}{3}-2 \frac{1}{2}\) \(B=1 \frac{3}{4}+1 \frac{1}{6}\)

Hints

- Calculate \(A\) and \(B\) separately first. - To compare fractional parts, rewrite them with a common denominator. - When subtracting mixed numbers, check whether you need to regroup one whole.

Solution

1. Calculate \(A\): \(5 \frac{1}{3}-2 \frac{1}{2}=5 \frac{2}{6}-2 \frac{3}{6}=4 \frac{8}{6}-2 \frac{3}{6}=2 \frac{5}{6}\). 2. Calculate \(B\): \(1 \frac{3}{4}+1 \frac{1}{6}=1 \frac{9}{12}+1 \frac{2}{12}=2 \frac{11}{12}\). 3. Rewrite \(A\) with denominator \(12\): \(2 \frac{5}{6}=2 \frac{10}{12}\). Since \(2 \frac{10}{12}<2 \frac{11}{12}\), \(A<B\).

Answer

\(A<B\), because \(A=2 \frac{5}{6}=2 \frac{10}{12}\) and \(B=2 \frac{11}{12}\).
5106307
Evaluate the expression using the correct order of operations. Write the answer as a mixed number. \(7 \frac{1}{5}-\left(2 \frac{1}{2}+1 \frac{3}{4}\right)+0.8\)

Hints

- Evaluate the expression inside the parentheses first. - Before changing the order of terms, rewrite subtraction in a form that can be regrouped using addition properties. - You can rewrite the decimal as a fraction before calculating.

Solution

1. Evaluate the parentheses: \(2 \frac{1}{2}+1 \frac{3}{4}=2 \frac{2}{4}+1 \frac{3}{4}=4 \frac{1}{4}\). 2. Rewrite \(0.8\) as \(\frac{4}{5}\), and rewrite subtraction as addition of the opposite: \(7 \frac{1}{5}+\left(-4 \frac{1}{4}\right)+\frac{4}{5}\). 3. Use the commutative and associative properties of addition to group the fifths: \(\left(7 \frac{1}{5}+\frac{4}{5}\right)+\left(-4 \frac{1}{4}\right)=8-4 \frac{1}{4}\). 4. Subtract: \(8-4 \frac{1}{4}=3 \frac{3}{4}\).

Answer

\(3 \frac{3}{4}\)
5106537
Simplify and remove the parentheses, then write an equivalent expression whose grouping makes the final calculation efficient. Evaluate the expression. \(\left(6\frac{4}{15} + \frac{18}{24}\right) - \left(1\frac{4}{15} - \frac{1}{4}\right)\)

Hints

- Reduce any fraction that is not in simplest form. - When a minus sign is in front of parentheses, track the sign of every term inside. - Look for an equivalent grouping that creates whole numbers.

Solution

1. Simplify \(\frac{18}{24}=\frac{3}{4}\) and distribute the subtraction: \(6\frac{4}{15}+\frac{3}{4}-1\frac{4}{15}+\frac{1}{4}\). 2. Regroup: \(\left(6\frac{4}{15}-1\frac{4}{15}\right)+\left(\frac{3}{4}+\frac{1}{4}\right)\). 3. The groups equal \(5\) and \(1\), so the value is \(6\).

Answer

After simplifying and removing parentheses: \(6\frac{4}{15}+\frac{3}{4}-1\frac{4}{15}+\frac{1}{4}\). One efficient regrouping is \(\left(6\frac{4}{15}-1\frac{4}{15}\right)+\left(\frac{3}{4}+\frac{1}{4}\right)=6\).
5106547
Rewrite the subtraction as addition of opposites. Then show an equivalent rearranged and regrouped sum that makes the calculation efficient, evaluate it, and name the properties that justify changing the order and grouping. \(\left(15\frac{5}{6} + 3\frac{1}{4}\right) - \left(2\frac{5}{6} + 1\frac{1}{2} + 1\frac{3}{4}\right)\)

Hints

- Rewrite each subtraction as addition of the opposite before changing the order. - Keep each sign attached to its term when rearranging. - Look for groups whose fractional parts cancel or combine to whole numbers.

Solution

1. Rewrite subtraction as addition of opposites: \(15\frac{5}{6}+3\frac{1}{4}+\left(-2\frac{5}{6}\right)+\left(-1\frac{1}{2}\right)+\left(-1\frac{3}{4}\right)\). 2. Use the commutative and associative properties of addition to regroup: \(\left(15\frac{5}{6}-2\frac{5}{6}\right)+\left(3\frac{1}{4}-1\frac{1}{2}-1\frac{3}{4}\right)\). 3. The first group is \(13\), and the second group is \(0\). 4. Therefore, the value is \(13\).

Answer

One valid regrouping is \(\left(15\frac{5}{6}-2\frac{5}{6}\right)+\left(3\frac{1}{4}-1\frac{1}{2}-1\frac{3}{4}\right)=13+0=13\). The commutative and associative properties of addition justify the rearrangement and regrouping after subtraction is rewritten as addition of opposites.
5106627
Evaluate each expression. Before calculating, state which representation you will use for each nonmatching pair and show the first rewritten expression. a) \(\left(\frac{1}{4} + 0.5\right) - 1\) b) \(2.75 - \left(1\frac{1}{2} + 0.25\right)\) c) \(\frac{1}{3} + 0.2 + \frac{1}{6}\)

Hints

- Inspect whether a fraction terminates before deciding to convert it. - A useful representation should reduce, not increase, the amount of computation. - Your answer must show the rewritten expression that reflects your choice.

Solution

1. a) Use decimals: \((0.25+0.5)-1=-0.25\). 2. b) Use decimals: \(2.75-(1.5+0.25)=1\). 3. c) Keep the thirds and sixths as fractions first: \(\left(\frac{1}{3}+\frac{1}{6}\right)+0.2=0.5+0.2=0.7\).

Answer

a) Decimals: \((0.25+0.5)-1=-0.25\) b) Decimals: \(2.75-(1.5+0.25)=1\) c) Fractions first: \(\left(\frac{1}{3}+\frac{1}{6}\right)+0.2=0.7\)
5106897
Evaluate the expression. Briefly explain why fractions or decimals are the better representation. \(1\frac{1}{3} + 0.25 - \frac{7}{12}\)

Hints

- Consider the decimal representation of \(\frac{1}{3}\). - Convert the terminating decimal to a fraction. - Rewrite the fractions with a common denominator.

Solution

1. Fractions are more efficient because \(\frac{1}{3}\) has a repeating decimal representation. 2. Convert \(0.25 = \frac{1}{4}\) and \(1\frac{1}{3} = \frac{4}{3}\). 3. Use denominator \(12\): \(\frac{16}{12} + \frac{3}{12} - \frac{7}{12} = \frac{12}{12} = 1\).

Answer

\(1\). Fractions are preferable because \(\frac{1}{3}\) is a repeating decimal.
5110997
Consider \(\frac{3}{4}+\frac{2}{3}-\frac{1}{2}\). a) Give a benchmark estimate that decides whether the result is greater than or less than \(1\). Show the benchmark values you used. b) Calculate the exact value as a fraction. c) State whether the exact result is consistent with your estimate.

Hints

- Use simple benchmark decimal values rather than doing the exact fraction computation first. - For the exact calculation, choose a common denominator for \(2\), \(3\), and \(4\). - Compare the exact result with the claim made by your estimate.

Solution

1. A benchmark estimate is \(0.75+0.67-0.5\approx0.92\), so the result should be less than \(1\). 2. With denominator \(12\), the exact value is \(\frac{9}{12}+\frac{8}{12}-\frac{6}{12}=\frac{11}{12}\). 3. Since \(\frac{11}{12}<1\), the estimate is consistent.

Answer

a) For example, \(0.75+0.67-0.5\approx0.92<1\). b) \(\frac{11}{12}\) c) Yes. The exact value is less than \(1\), as predicted.
5111867
Which expression has the greater value? Evaluate both expressions and justify your answer. Expression A: \(\left(\frac{3}{4}-1.25\right)\cdot(-4)\) Expression B: \(2.5\div\frac{1}{2}-6\)

Hints

- Evaluate the two expressions separately. - It may help to write fractions and decimals in the same form. - How do you divide by a fraction?

Solution

1. Expression A: \(\frac{3}{4}=0.75\), so \(0.75-1.25=-0.5\). Then \((-0.5)\cdot(-4)=2\). 2. Expression B: \(2.5\div\frac{1}{2}=5\), so \(5-6=-1\). 3. Since \(2>-1\), Expression A has the greater value.

Answer

Expression A is greater because \(2>-1\).
5113167
Check the following work for an error: \(\frac{4}{5}-\left(\frac{1}{2}-\frac{1}{10}\right)=\frac{4}{5}-\frac{1}{2}-\frac{1}{10}=\frac{8}{10}-\frac{5}{10}-\frac{1}{10}=\frac{1}{5}\). Explain whether the work is correct. If it is not, identify the error and calculate the correct result.

Hints

- Try evaluating the expression inside the parentheses before doing anything else. - What quantity is being subtracted from \(\frac{4}{5}\)? - Check each equality in the student's chain.

Solution

1. The work is incorrect. Subtracting the entire quantity \(\frac{1}{2}-\frac{1}{10}\) is not the same as subtracting both terms separately. 2. Evaluate the parentheses first: \(\frac{1}{2}-\frac{1}{10}=\frac{5}{10}-\frac{1}{10}=\frac{4}{10}\). 3. Then \(\frac{4}{5}-\frac{4}{10}=\frac{8}{10}-\frac{4}{10}=\frac{4}{10}=\frac{2}{5}\).

Answer

The work is incorrect because the subtraction of the quantity in parentheses was handled incorrectly. The correct result is \(\frac{2}{5}\).
5113387
Evaluate the expression: \(\left(\frac{2}{3}-1.5\right)+\left(-\frac{5}{6}+0.25\right)\)

Hints

- Decide whether fractions or decimals will be easier to use consistently. - Fractions need common denominators before addition or subtraction. - Evaluate the expressions inside parentheses first.

Solution

1. First parentheses: \(\frac{2}{3}-1.5=\frac{2}{3}-\frac{3}{2}=\frac{4}{6}-\frac{9}{6}=-\frac{5}{6}\). 2. Second parentheses: \(-\frac{5}{6}+0.25=-\frac{5}{6}+\frac{1}{4}=-\frac{10}{12}+\frac{3}{12}=-\frac{7}{12}\). 3. Add: \(-\frac{5}{6}-\frac{7}{12}=-\frac{10}{12}-\frac{7}{12}=-\frac{17}{12}=-1 \frac{5}{12}\).

Answer

\(-\frac{17}{12}\), or \(-1 \frac{5}{12}\)
5113397
Evaluate the expression: \(\left(-4.5\div\frac{9}{10}\right)\cdot\left(\frac{2}{3}-1.2\right)\)

Hints

- Apply the order of operations inside each set of parentheses. - How do you divide by a fraction? - Simplify the final fraction or write it as a mixed number.

Solution

1. Evaluate the first parentheses: \(-4.5=-\frac{9}{2}\), so \(-\frac{9}{2}\div\frac{9}{10}=-\frac{9}{2}\cdot\frac{10}{9}=-5\). 2. Evaluate the second parentheses: \(1.2=\frac{6}{5}\), so \(\frac{2}{3}-\frac{6}{5}=\frac{10}{15}-\frac{18}{15}=-\frac{8}{15}\). 3. Multiply: \(-5\cdot\left(-\frac{8}{15}\right)=\frac{8}{3}=2 \frac{2}{3}\).

Answer

\(\frac{8}{3}\), or \(2 \frac{2}{3}\)
5113577
Evaluate the expression: \(\frac{5}{8}-\left[(-2.5)\cdot\left(\frac{1}{5}-0.6\right)\right]\div\left(-\frac{1}{2}\right)\)

Hints

- Start with the innermost parentheses. - What sign results when two negative numbers are multiplied? - How do you divide by a fraction? - Decide whether a fraction or decimal form is more useful for the final result.

Solution

1. Evaluate the inner parentheses: \(\frac{1}{5}-0.6=0.2-0.6=-0.4\). 2. Multiply inside the brackets: \((-2.5)\cdot(-0.4)=1\). 3. Divide: \(1\div\left(-\frac{1}{2}\right)=-2\). 4. Subtract: \(\frac{5}{8}-(-2)=\frac{21}{8}=2 \frac{5}{8}=2.625\).

Answer

\(2 \frac{5}{8}\), or \(2.625\)
5113767
Two students evaluate \(4.8+2\frac{1}{4}+1.2+\frac{3}{4}\). - Jordan converts the fractions to decimals and adds from left to right. - Maya reorders and groups the addends as \((4.8+1.2)+(2\frac{1}{4}+\frac{3}{4})\). a) Evaluate the expression both ways. b) Explain which method is easier to do mentally.

Hints

- Complete both methods before comparing them. - Look for addends that combine to whole numbers. - Consider how reordering and regrouping affect the value of a sum.

Solution

1. Jordan’s method: \(2\frac{1}{4}=2.25\) and \(\frac{3}{4}=0.75\). Then \(4.8+2.25=7.05\), \(7.05+1.2=8.25\), and \(8.25+0.75=9\). 2. Maya’s method: \(4.8+1.2=6\) and \(2\frac{1}{4}+\frac{3}{4}=3\). Then \(6+3=9\). 3. Maya’s method is easier mentally because the commutative and associative properties create whole-number sums.

Answer

a) Both methods give \(9\). b) Maya’s method is generally easier mentally because the grouped addends make \(6\) and \(3\).
5117117
Decide whether each statement is true or false. Justify your decision by evaluating both sides of the equation. a) \(-12.5+7.5=-(12.5-7.5)\) b) \(\frac{1}{2}-\left(\frac{1}{4}+\frac{1}{8}\right)=\frac{1}{2}-\frac{1}{4}+\frac{1}{8}\)

Hints

- Evaluate each side separately. - Apply parentheses first. - What happens to every term inside parentheses when a negative sign is distributed?

Solution

1. For a), the left side is \(-12.5+7.5=-5\). The right side is \(-(12.5-7.5)=-5\). The statement is true. 2. For b), the left side is \(\frac{1}{2}-\left(\frac{1}{4}+\frac{1}{8}\right)=\frac{4}{8}-\frac{3}{8}=\frac{1}{8}\). 3. The right side is \(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}=\frac{4}{8}-\frac{2}{8}+\frac{1}{8}=\frac{3}{8}\). Since the two sides differ, the statement is false.

Answer

a) True; both sides equal \(-5\). b) False; the left side is \(\frac{1}{8}\) and the right side is \(\frac{3}{8}\).
5117967
Evaluate the expression. Write the result as a fraction in simplest form or as a mixed number: \(10-\left(3 \frac{1}{4}+2 \frac{5}{6}\right)-1 \frac{1}{12}\)

Hints

- Evaluate the parentheses first. - When subtracting a mixed number from a whole number, you may need to regroup one whole. - Simplify the fractional part of the final answer. - After the parentheses are resolved, keep the remaining subtractions in their original order.

Solution

1. Evaluate the parentheses using denominator \(12\): \(3 \frac{1}{4}+2 \frac{5}{6}=3 \frac{3}{12}+2 \frac{10}{12}=5 \frac{13}{12}=6 \frac{1}{12}\). 2. Subtract from \(10\): \(10-6 \frac{1}{12}=3 \frac{11}{12}\). 3. Subtract the last mixed number: \(3 \frac{11}{12}-1 \frac{1}{12}=2 \frac{10}{12}=2 \frac{5}{6}\).

Answer

\(2 \frac{5}{6}\)
5118147
Evaluate the expression. Write the answer as a fraction in simplest form or as a mixed number: \(2 \frac{1}{4}-\frac{5}{6}-\frac{3}{8}\)

Hints

- Rewrite the mixed number as an improper fraction. - Find one common denominator for all three fractions. - Perform the subtractions from left to right, then convert the result to a mixed number if needed. - Estimate whether the final value should be greater than \(1\) as a reasonableness check.

Solution

1. Rewrite \(2 \frac{1}{4}\) as \(\frac{9}{4}\). 2. Use denominator \(24\): \(\frac{9}{4}=\frac{54}{24}\), \(\frac{5}{6}=\frac{20}{24}\), and \(\frac{3}{8}=\frac{9}{24}\). 3. Subtract from left to right: \(\frac{54}{24}-\frac{20}{24}-\frac{9}{24}=\frac{25}{24}=1 \frac{1}{24}\).

Answer

\(1 \frac{1}{24}\)
5121627
Find the missing value of \(a\), \(b\), or \(a+b\) in each part. a) \(a=-4.5\), \(b=2.1\), \(a+b=\square\) b) \(a=\square\), \(b=-7.8\), \(a+b=-3.2\) c) \(a=\frac{3}{4}\), \(b=\square\), \(a+b=-\frac{1}{8}\)

Hints

- When the sum and one addend are known, subtract the known addend to find the other one. - Subtraction is the inverse of addition. - For the fractions, first write them with a common denominator.

Solution

1. In a), add the two values: \(-4.5+2.1=-2.4\). 2. In b), subtract the known addend from the sum: \(a=-3.2-(-7.8)=-3.2+7.8=4.6\). 3. In c), subtract the known addend from the sum: \(b=-\frac{1}{8}-\frac{3}{4}=-\frac{1}{8}-\frac{6}{8}=-\frac{7}{8}\).

Answer

a) \(a+b=-2.4\) b) \(a=4.6\) c) \(b=-\frac{7}{8}\)
5121757
Complete the subtraction table and find \(x\) and \(y\). Each entry is the value at the left of its row minus the value at the top of its column. <table> <tr><td>\(-\)</td><td>\(1.2\)</td><td>\(y\)</td><td>\(-3.5\)</td></tr> <tr><td>\(4.5\)</td><td>\(3.3\)</td><td>\(7.2\)</td><td></td></tr> <tr><td>\(-2.8\)</td><td></td><td>\(-0.1\)</td><td>\(0.7\)</td></tr> <tr><td>\(x\)</td><td>\(-5.5\)</td><td></td><td>\(-0.8\)</td></tr> </table>

Hints

- First determine how each table entry is calculated. - Use a simple equation to find a missing row or column heading. - Pay close attention when subtracting a negative number.

Solution

1. From the row headed by \(4.5\), \(4.5-y=7.2\). Therefore, \(y=4.5-7.2=-2.7\). 2. From the row headed by \(x\), \(x-1.2=-5.5\). Therefore, \(x=-5.5+1.2=-4.3\). 3. Fill the remaining cells: \(4.5-(-3.5)=8.0\), \(-2.8-1.2=-4.0\), and \(-4.3-(-2.7)=-1.6\).

Answer

\(x=-4.3\); \(y=-2.7\) Completed table: <table> <tr><td>\(-\)</td><td>\(1.2\)</td><td>\(-2.7\)</td><td>\(-3.5\)</td></tr> <tr><td>\(4.5\)</td><td>\(3.3\)</td><td>\(7.2\)</td><td>\(8.0\)</td></tr> <tr><td>\(-2.8\)</td><td>\(-4.0\)</td><td>\(-0.1\)</td><td>\(0.7\)</td></tr> <tr><td>\(-4.3\)</td><td>\(-5.5\)</td><td>\(-1.6\)</td><td>\(-0.8\)</td></tr> </table>
5121797
Insert \(<\), \(>\), or \(=\) to make each statement true. For each part, justify the comparison using a rational-number rule or magnitude comparison without giving the exact value of the expression. a) \(-4.5+4.5\;\dots\;0\) b) \(-123-(-122)\;\dots\;0\) c) \(-\frac{2}{3}+\left(-\frac{1}{3}\right)\;\dots\;0\) d) \(0.01-0.1\;\dots\;0\)

Hints

- Use structural facts such as opposites and sign rules before doing arithmetic. - When signs differ, compare magnitudes. - Your explanation should establish the comparison with \(0\), not report an exact result.

Solution

1. a) Opposites sum to \(0\), so use \(=\). 2. b) Rewrite as \(-123+122\). The negative magnitude is larger, so the result is less than \(0\). 3. c) A sum of two negative numbers is negative, so use \(<\). 4. d) The larger positive number is subtracted from the smaller, so the result is negative and use \(<\).

Answer

a) \(=\) — the addends are opposites b) \(<\) — \(-123+122\) has the larger magnitude on the negative term c) \(<\) — two negative addends have a negative sum d) \(<\) — \(0.1>0.01\), so the difference is negative
5121807
For each pair, decide which expression has the greater value. Justify the choice using signs, direction on a number line, or relative magnitude; do not give the exact final values. a) \((-15)+(-20)\) or \((-15)-(-20)\) b) \(0.4-0.6\) or \(-0.4+0.6\) c) \(-\frac{1}{8}-\frac{1}{4}\) or \(-\frac{1}{8}+\frac{1}{4}\)

Hints

- Look for a sign comparison that settles the question before exact calculation. - Rewrite subtraction of a negative as addition. - For the fraction pair, think about movement from the same starting point.

Solution

1. a) The first expression is a sum of negatives. The second rewrites as \(-15+20\), which is positive, so the second is greater. 2. b) Subtracting a larger positive from a smaller gives a negative result, while adding \(0.6\) to \(-0.4\) gives a positive result, so the second is greater. 3. c) From \(-\frac{1}{8}\), subtracting \(\frac{1}{4}\) moves left while adding it moves right, so the second is greater.

Answer

a) \((-15)-(-20)\) — it is positive while the other expression is negative b) \(-0.4+0.6\) — it is positive while \(0.4-0.6\) is negative c) \(-\frac{1}{8}+\frac{1}{4}\) — adding the positive fraction moves right while subtracting it moves left
5121857
Let \(x=-5\) and \(y=-2\). For each expression, first predict whether the result is positive or negative. Then find the exact value. a) \(xy\) b) \(x+y\) c) \(y-x\) d) \(\frac{x}{y}\)

Hints

- Review the sign rules for multiplication and division. - Adding two negative numbers gives a negative sum. - Subtracting a negative number is equivalent to adding.

Solution

1. The product of two negative numbers is positive: \((-5)\cdot(-2)=10\). 2. The sum of two negative numbers is negative: \(-5+(-2)=-7\). 3. Subtracting a negative number gives \(y-x=-2-(-5)=-2+5=3\), which is positive. 4. The quotient of two negative numbers is positive: \(\frac{-5}{-2}=\frac{5}{2}=2.5\).

Answer

a) Positive; \(10\) b) Negative; \(-7\) c) Positive; \(3\) d) Positive; \(2.5\)
5121967
Choose exactly three different numbers from the list whose sum is \(0\). Find two different solutions. \(0.7,\ -0.3,\ \frac{1}{5},\ -0.4,\ 0.1,\ -0.5\)

Hints

- Convert the fraction to a decimal. - For each positive number, look for two numbers whose sum is its opposite. - Use each selected number only once in a solution.

Solution

1. Write \(\frac{1}{5}=0.2\). 2. One solution is \(0.7+(-0.3)+(-0.4)=0\). 3. A second solution is \(0.2+0.1+(-0.3)=0\).

Answer

For example: \(0.7, -0.3, -0.4\) \(\frac{1}{5}, 0.1, -0.3\)
5121977
Choose exactly four numbers from the list. Place a plus or minus sign between consecutive chosen numbers so that the result is \(0.5\). \(1.2,\ -0.5,\ \frac{1}{4},\ -0.8,\ 0.1,\ -0.2\)

Hints

- Convert \(\frac{1}{4}\) to a decimal if you use it. - Estimate combinations before calculating exactly. - Remember that subtracting a negative number adds its opposite.

Solution

1. One valid choice is \(1.2, -0.8, 0.1, -0.2\). 2. Insert signs: \(1.2+(-0.8)-0.1-(-0.2)\). 3. Evaluate: \(1.2-0.8-0.1+0.2=0.5\).

Answer

One possible solution is \(1.2+(-0.8)-0.1-(-0.2)=0.5\).
5122397
Two students evaluate \(-0.5 + \frac{3}{4} - 0.25 + \frac{1}{2}\) in different ways. Mia works from left to right. Lucas groups values that combine easily. a) Evaluate the expression using Mia’s method. b) Show an efficient grouping Lucas could use. c) Compare the original expression with \(-0.5 + \left(\frac{3}{4} - 0.25\right) + \frac{1}{2}\). Do the parentheses change the value? Explain.

Hints

- Convert a fraction to a decimal only when it helps with a particular step. - Look for additive inverses or pairs that make a simple decimal. - Parentheses preceded by a plus sign can be removed without changing the signs inside.

Solution

1. Mia’s method: \(-0.5 + 0.75 = 0.25\), then \(0.25 - 0.25 = 0\), and finally \(0 + 0.5 = 0.5\). 2. Lucas can group \(-0.5\) with \(\frac{1}{2}\), and \(\frac{3}{4}\) with \(-0.25\): \(\left(-0.5 + \frac{1}{2}\right) + \left(\frac{3}{4} - 0.25\right)\). 3. The groups equal \(0\) and \(0.5\), so the result is \(0.5\). 4. In c), the parentheses are preceded by addition. Removing them does not change any signs, so the value remains \(0.5\).

Answer

a) \(0.5\) b) \(\left(-0.5 + \frac{1}{2}\right) + \left(\frac{3}{4} - 0.25\right) = 0 + 0.5 = 0.5\) c) No. The parentheses are preceded by addition, so removing them does not change the value.
5122427
For each expression, write an equivalent grouping that combines compatible rational numbers before evaluating. a) \(-\frac{3}{7} + \frac{5}{8} - \frac{4}{7} + \frac{3}{8}\) b) \(1\frac{2}{5} - 3\frac{1}{4} + \frac{3}{5} + \frac{1}{4}\) c) \(\frac{7}{10} - 0.45 + \frac{3}{10} - 0.55\)

Hints

- Identify which terms can combine without introducing new denominators or representations. - Treat subtraction as addition of a signed term before regrouping. - Show the grouping that makes each calculation short.

Solution

1. a) \(\left(-\frac{3}{7}-\frac{4}{7}\right)+\left(\frac{5}{8}+\frac{3}{8}\right)=-1+1=0\). 2. b) \(\left(1\frac{2}{5}+\frac{3}{5}\right)+\left(-3\frac{1}{4}+\frac{1}{4}\right)=2-3=-1\). 3. c) \(\left(\frac{7}{10}+\frac{3}{10}\right)+(-0.45-0.55)=1-1=0\).

Answer

a) \(\left(-\frac{3}{7}-\frac{4}{7}\right)+\left(\frac{5}{8}+\frac{3}{8}\right)=0\) b) \(\left(1\frac{2}{5}+\frac{3}{5}\right)+\left(-3\frac{1}{4}+\frac{1}{4}\right)=-1\) c) \(\left(\frac{7}{10}+\frac{3}{10}\right)+(-0.45-0.55)=0\)
5122437
Lina and Tim evaluate \(12.5 - 8.3 + 7.5 - 1.7\) in different ways. Lina groups the positive terms and the amounts being subtracted: \((12.5 + 7.5) - (8.3 + 1.7)\). Tim groups consecutive pairs: \((12.5 - 8.3) + (7.5 - 1.7)\). a) Evaluate the expression using both methods. b) Explain why Lina’s method is especially efficient here.

Hints

- Carry out each proposed grouping separately. - Compare the intermediate values produced by the two methods. - An efficient mental method often creates whole numbers.

Solution

1. Lina’s method: \(12.5 + 7.5 = 20\) and \(8.3 + 1.7 = 10\), so \(20 - 10 = 10\). 2. Tim’s method: \(12.5 - 8.3 = 4.2\) and \(7.5 - 1.7 = 5.8\), so \(4.2 + 5.8 = 10\). 3. Lina’s grouping creates whole-number intermediate results, making the mental calculation simpler.

Answer

a) Both methods give \(10\). b) Lina’s method creates the whole-number intermediate results \(20\) and \(10\).
5122457
Evaluate each expression using properties of rational-number addition. For each part, show the reordered or regrouped expression that makes the calculation efficient. a) \(\frac{5}{9} - \frac{1}{3} + \frac{4}{9}\) b) \(\frac{3}{4} - 2.5 - \frac{7}{4} + 1.5\) c) \(-\frac{2}{7} + \frac{5}{6} + \frac{2}{7} - \frac{1}{6}\)

Hints

- Search for equal denominators, additive inverses, or decimal pairs that make whole numbers. - Keep signs attached to terms while changing order. - The answer must show how the properties were used.

Solution

1. a) \(\left(\frac{5}{9}+\frac{4}{9}\right)-\frac{1}{3}=1-\frac{1}{3}=\frac{2}{3}\). 2. b) \(\left(\frac{3}{4}-\frac{7}{4}\right)+(-2.5+1.5)=-1-1=-2\). 3. c) \(\left(-\frac{2}{7}+\frac{2}{7}\right)+\left(\frac{5}{6}-\frac{1}{6}\right)=\frac{2}{3}\).

Answer

a) \(\left(\frac{5}{9}+\frac{4}{9}\right)-\frac{1}{3}=\frac{2}{3}\) b) \(\left(\frac{3}{4}-\frac{7}{4}\right)+(-2.5+1.5)=-2\) c) \(\left(-\frac{2}{7}+\frac{2}{7}\right)+\left(\frac{5}{6}-\frac{1}{6}\right)=\frac{2}{3}\)
5122467
Rewrite each expression by removing parentheses or regrouping strategically, then evaluate. a) \(-3.25-(1.75-8)\) b) \(2\frac{1}{5}-\left(4.8+\frac{1}{5}\right)+0.8\) c) \(\left(\frac{5}{8}+1.4\right)-\left(\frac{1}{8}-0.6\right)\)

Hints

- A subtraction sign before parentheses changes the sign of every term inside. - Convert a fraction to a decimal when that creates easy pairs. - Regroup terms to form whole numbers.

Solution

1. For a), remove the parentheses: \(-3.25-1.75+8=-5+8=3\). 2. For b), remove the parentheses and regroup: \(2\frac{1}{5}-4.8-\frac{1}{5}+0.8=(2.2-0.2)+(-4.8+0.8)=2-4=-2\). 3. For c), remove the parentheses: \(\frac{5}{8}+1.4-\frac{1}{8}+0.6\). Regroup: \(\left(\frac{5}{8}-\frac{1}{8}\right)+(1.4+0.6)=0.5+2=2.5\).

Answer

a) \(3\) b) \(-2\) c) \(2.5\)
5122657
Use all four number cards \(-4\), \(-2\), \(6\), and \(8\), along with addition and subtraction signs, to write an expression with a value of \(0\). You may use parentheses. Then write a different expression using the same four cards that has a value of \(12\).

Hints

- First consider how the absolute values \(4\), \(2\), \(6\), and \(8\) can combine to make each target. - Subtracting a negative number is the same as adding its opposite. - It may help to consider the positive and negative cards separately.

Solution

1. One expression with a value of \(0\) is \(8-6-(-2)+(-4)\). Evaluating gives \(8-6+2-4=0\). 2. One expression with a value of \(12\) is \(8+6-(-2)+(-4)\). Evaluating gives \(8+6+2-4=12\).

Answer

For \(0\): \(8-6-(-2)+(-4)=0\) For \(12\): \(8+6-(-2)+(-4)=12\)
5122677
Insert an addition or subtraction sign in each blank to make the equation true. a) \(-7\;\square\;(-3)\;\square\;5=-9\) b) \(-7\;\square\;((-3)\;\square\;5)=1\)

Hints

- Test the four possible pairs of operation signs systematically. - In part b), evaluate the expression in parentheses first. - Decide whether the result must be greater than or less than the starting value \(-7\).

Solution

1. For a), using subtraction in both blanks gives \(-7-(-3)-5=-7+3-5=-9\). 2. For b), evaluate the parentheses first. Using subtraction in both blanks gives \(-7-((-3)-5)=-7-(-8)=1\).

Answer

a) \(-7-(-3)-5=-9\) b) \(-7-((-3)-5)=1\)
5122847
A point is at \(-3.4\) on a number line. A second point at coordinate \(x\) is exactly \(7.2\) units away. a) Write an absolute-value equation that models the distance condition. b) Solve the equation by considering the two possible signed differences.

Hints

- Model “distance from \(-3.4\)” with an absolute difference involving \(x\). - An absolute value equal to a positive number produces two signed cases. - Check that both solutions are exactly \(7.2\) units from \(-3.4\).

Solution

1. The distance equation is \(|x-(-3.4)|=7.2\), or \(|x+3.4|=7.2\). 2. Thus, \(x+3.4=7.2\) or \(x+3.4=-7.2\). 3. The solutions are \(x=3.8\) and \(x=-10.6\).

Answer

a) \(|x+3.4|=7.2\) b) \(x=3.8\) or \(x=-10.6\)
5122857
For each pair, write an absolute-difference expression and evaluate the distance. Give each answer as a decimal or a fraction in simplest form. a) \(-\frac{3}{4}\) and \(0.2\) b) \(1\frac{1}{2}\) and \(-\frac{2}{5}\)

Hints

- Use an absolute difference before converting to a convenient common representation. - The absolute value ensures the distance is nonnegative regardless of subtraction order. - Keep the absolute-value expression in the final response.

Solution

1. a) \(0.2=\frac{1}{5}\), so \(\left|\frac{1}{5}-\left(-\frac{3}{4}\right)\right|=\frac{19}{20}=0.95\). 2. b) \(1\frac{1}{2}=\frac{3}{2}\), so \(\left|\frac{3}{2}-\left(-\frac{2}{5}\right)\right|=\frac{19}{10}=1.9\).

Answer

a) \(\left|0.2-\left(-\frac{3}{4}\right)\right|=0.95=\frac{19}{20}\) b) \(\left|1\frac{1}{2}-\left(-\frac{2}{5}\right)\right|=1.9=\frac{19}{10}\)
5122997
A number line is drawn so that \(1\) unit is \(4\,\text{cm}\). a) How many centimeters from \(0\) should \(-1.25\) be placed, and in which direction? b) What number is exactly halfway between \(-\frac{1}{2}\) and \(\frac{3}{4}\)? Give the answer as a fraction in simplest form and as a decimal.

Hints

- Multiply the numerical distance from \(0\) by the scale in centimeters per unit. - The midpoint is the average of the two endpoints. - Use a common denominator before adding the fractions.

Solution

1. For a), the distance from \(-1.25\) to \(0\) is \(1.25\) units. The drawing distance is \(1.25\cdot4\,\text{cm}=5\,\text{cm}\), to the left of \(0\). 2. For b), find the midpoint: \(\frac{-\frac{1}{2}+\frac{3}{4}}{2}=\frac{\frac{1}{4}}{2}=\frac{1}{8}\). 3. As a decimal, \(\frac{1}{8}=0.125\).

Answer

a) \(5\,\text{cm}\) to the left of \(0\) b) \(\frac{1}{8}=0.125\)
5123027
Evaluate each expression using properties of operations. For each part, show the cancellation or regrouping that makes the calculation short. a) \(17.4 - (5.9 + 17.4) + 0.9\) b) \(-\frac{2}{5} + (3.7 + 0.4) - 2.7\)

Hints

- First rewrite any subtraction of a grouped sum carefully. - Look for additive inverses or pairs differing by a whole number. - Show the cancellation or regrouping, not only the result.

Solution

1. a) Remove the parentheses: \(17.4-5.9-17.4+0.9\). Regroup as \((17.4-17.4)+(-5.9+0.9)=-5\). 2. b) Convert \(-\frac{2}{5}=-0.4\). Then regroup as \((-0.4+0.4)+(3.7-2.7)=1\).

Answer

a) \((17.4-17.4)+(-5.9+0.9)=-5\) b) \((-0.4+0.4)+(3.7-2.7)=1\)
5123037
Rewrite and evaluate each expression. Pay attention to the mixture of fractions and decimals. a) \(\frac{1}{8}-0.25+0.875-1.75\) b) \(\left(\frac{5}{6}+1.2\right)-\left(\frac{1}{6}-0.8\right)\)

Hints

- Choose fraction or decimal forms that make useful pairs. - Combine fractions with the same denominator directly. - Simplify the final fraction.

Solution

1. For a), write \(\frac{1}{8}=0.125\). Regroup: \((0.125+0.875)-(0.25+1.75)=1-2=-1\). 2. For b), remove the parentheses: \(\frac{5}{6}+1.2-\frac{1}{6}+0.8\). 3. Regroup: \(\left(\frac{5}{6}-\frac{1}{6}\right)+(1.2+0.8)=\frac{4}{6}+2=\frac{2}{3}+2=2\frac{2}{3}\).

Answer

a) \(-1\) b) \(2\frac{2}{3}\), or \(\frac{8}{3}\)
5124697
Place exactly one pair of parentheses in \(40-20-10-5\). The parentheses must contain at least two numbers. a) Find the value without parentheses. b) Place the parentheses so the value does not change. c) Place the parentheses so the value is \(25\).

Hints

- Without parentheses, perform subtraction from left to right. - A minus sign before parentheses changes how the enclosed difference affects the result. - Test possible groups of consecutive numbers.

Solution

1. Without parentheses, calculate from left to right: \(40-20-10-5=5\). 2. One placement that keeps the value is \((40-20)-10-5=5\). Another is \((40-20-10)-5=5\). 3. To obtain \(25\), use \(40-(20-10)-5=25\).

Answer

a) \(5\) b) \((40-20)-10-5\) or \((40-20-10)-5\) c) \(40-(20-10)-5\)
5124707
Consider the expression \(50-25-10-5\). Determine whether placing exactly one pair of parentheses can make the value \(40\). The parentheses must enclose at least two numbers. Justify your answer by evaluating every possible placement.

Hints

- Count all the ways one pair of parentheses can enclose consecutive numbers in a four-number expression. - Evaluate each possible placement carefully. - Remember to simplify the expression inside parentheses first.

Solution

1. Parentheses around the first two numbers give \((50-25)-10-5=10\). 2. Parentheses around the first three numbers give \((50-25-10)-5=10\). 3. Parentheses around the entire expression give \((50-25-10-5)=10\). 4. Parentheses around the second and third numbers give \(50-(25-10)-5=30\). 5. Parentheses around the second through fourth numbers give \(50-(25-10-5)=40\). 6. Parentheses around the third and fourth numbers give \(50-25-(10-5)=20\). 7. Therefore, it is possible. The required placement is \(50-(25-10-5)\).

Answer

Yes. The placement \(50-(25-10-5)\) gives \(40\).
5128177
Evaluate the expression using the order of operations: \(18\div(-3)-4\cdot(-2.5)\)

Hints

- Multiplication and division come before subtraction. - What sign results when a positive number is divided by a negative number? - Pay attention to both the subtraction sign and the negative product.

Solution

1. Divide: \(18\div(-3)=-6\). 2. Multiply: \(4\cdot(-2.5)=-10\). 3. Subtract: \(-6-(-10)=4\).

Answer

\(4\)
5128207
A weather station starts the day at \(45.5\,{}^\circ\text{F}\). The temperature then changes by \(-3.4\,{}^\circ\text{F}\), \(+8.8\,{}^\circ\text{F}\), \(-2.2\,{}^\circ\text{F}\), \(-5.7\,{}^\circ\text{F}\), and \(+6.1\,{}^\circ\text{F}\), in that order. a) What is the final temperature? b) What is the highest temperature reached during the day? c) What is the lowest temperature reached during the day?

Hints

- Make a running list of the temperature after each change. - Include the starting temperature when finding the highest and lowest values. - Pay attention to the signs of the changes.

Solution

1. Track the temperatures in order: \(45.5\), \(42.1\), \(50.9\), \(48.7\), \(43.0\), and \(49.1\), all in degrees Fahrenheit. 2. For a), the final temperature is \(49.1\,{}^\circ\text{F}\). 3. For b), the highest value is \(50.9\,{}^\circ\text{F}\). 4. For c), the lowest value is \(42.1\,{}^\circ\text{F}\).

Answer

a) \(49.1\,{}^\circ\text{F}\) b) \(50.9\,{}^\circ\text{F}\) c) \(42.1\,{}^\circ\text{F}\)
5128217
A rain barrel holds \(50\) gallons. It starts the week with \(30.75\) gallons of water. During the week, the amount changes in this order: - Rain: \(+8.50\) gallons - Watering: \(-13.20\) gallons - Rain: \(+3.85\) gallons - Watering: \(-22.40\) gallons a) How much water is in the barrel at the end of the week? b) When the barrel reached its highest amount during the week, how many gallons of empty capacity remained?

Hints

- Update the amount after each addition or removal. - For part b), identify the greatest amount reached during the week. - Subtract that maximum amount from the barrel's total capacity.

Solution

1. Track the amounts: \(30.75\), \(39.25\), \(26.05\), \(29.90\), and \(7.50\) gallons. 2. For a), the final amount is \(7.50\) gallons. 3. The highest amount is \(39.25\) gallons. 4. For b), the empty capacity at that point is \(50.00-39.25=10.75\) gallons.

Answer

a) \(7.50\) gallons b) \(10.75\) gallons of empty capacity
5128227
A checking account starts with a balance of \(\$250.00\). These transactions occur in order: - Withdrawal: \(\$315.50\) - Deposit: \(\$120.00\) - Withdrawal: \(\$85.25\) a) What is the lowest balance reached during the sequence? b) What is the balance after the final transaction? c) How much must be deposited after the final transaction to bring the balance to exactly \(\$100.00\)?

Hints

- Update the account balance after each transaction in order. - A withdrawal decreases the balance, and the balance can become negative. - To move from a negative final balance to a positive target, find the difference between the two signed values.

Solution

1. Track the balances: \(\$250.00\), \(-\$65.50\), \(\$54.50\), and \(-\$30.75\). 2. For a), the lowest balance is \(-\$65.50\). 3. For b), the final balance is \(-\$30.75\). 4. For c), the required deposit is \(100.00-(-30.75)=130.75\), so \(\$130.75\) must be deposited.

Answer

a) \(-\$65.50\) b) \(-\$30.75\) c) \(\$130.75\)
5128347
Find the rational number that belongs in each blank. a) \(\frac{2}{3}+\square=-\frac{1}{6}\) b) \(\square-\frac{3}{5}=-1\) c) \(-\frac{1}{2}-\square=\frac{1}{8}\)

Hints

- Treat each blank as an unknown value and use the inverse operation. - Use the same inverse-operation reasoning you would use with positive numbers. - Rewrite rational numbers with common denominators before adding or subtracting. - In part c), pay close attention to the subtraction sign before the blank.

Solution

1. For a), the missing value is \(-\frac{1}{6}-\frac{2}{3}=-\frac{1}{6}-\frac{4}{6}=-\frac{5}{6}\). 2. For b), the missing value is \(-1+\frac{3}{5}=-\frac{5}{5}+\frac{3}{5}=-\frac{2}{5}\). 3. For c), the missing value is \(-\frac{1}{2}-\frac{1}{8}=-\frac{4}{8}-\frac{1}{8}=-\frac{5}{8}\).

Answer

a) \(-\frac{5}{6}\) b) \(-\frac{2}{5}\) c) \(-\frac{5}{8}\)
5142287
Write each expression and find its value. a) Decrease \(-18.5\) by the difference of \(12.4\) and \(15.9\). b) Add the opposite of \(7.2\) to the sum of \(-14.6\) and \(21.3\). c) Subtract three times \(4.5\) from the difference of \(-10\) and \(-25\).

Hints

- “Decrease by” indicates subtraction. - Find the opposite before adding in part b). - In “the difference of A and B,” calculate \(A-B\).

Solution

1. For a), \(-18.5-(12.4-15.9)=-18.5-(-3.5)=-15\). 2. For b), \((-14.6+21.3)+(-7.2)=6.7-7.2=-0.5\). 3. For c), \((-10-(-25))-3\cdot4.5=15-13.5=1.5\).

Answer

a) \(-15\) b) \(-0.5\) c) \(1.5\)
5142307
A weather station in Colorado records these temperature changes from one reading to the next. At midnight, the temperature is \(27.5\,{}^\circ\text{F}\). <table> <tr><td>6:00 a.m.</td><td>\(+7.2\,{}^\circ\text{F}\)</td></tr> <tr><td>9:00 a.m.</td><td>\(+6.8\,{}^\circ\text{F}\)</td></tr> <tr><td>12:00 p.m.</td><td>\(+4.4\,{}^\circ\text{F}\)</td></tr> <tr><td>3:00 p.m.</td><td>\(-2.5\,{}^\circ\text{F}\)</td></tr> <tr><td>6:00 p.m.</td><td>\(-8.6\,{}^\circ\text{F}\)</td></tr> <tr><td>9:00 p.m.</td><td>\(-4.3\,{}^\circ\text{F}\)</td></tr> </table> a) Write one expression for the temperature at 9:00 p.m. and calculate it. b) Find the highest and lowest temperatures at the listed reading times, including midnight.

Hints

- Start with the midnight temperature and apply each signed change in order. - Keep a list of the temperature after every change. - Include the starting value when finding the maximum and minimum. - A positive change raises the temperature; a negative change lowers it.

Solution

1. For a), use \(27.5+7.2+6.8+4.4-2.5-8.6-4.3=30.5\). The temperature at 9:00 p.m. is \(30.5\,{}^\circ\text{F}\). 2. The successive temperatures are \(27.5\), \(34.7\), \(41.5\), \(45.9\), \(43.4\), \(34.8\), and \(30.5\) degrees Fahrenheit. 3. For b), the highest temperature is \(45.9\,{}^\circ\text{F}\), and the lowest is \(27.5\,{}^\circ\text{F}\).

Answer

a) \(27.5+7.2+6.8+4.4-2.5-8.6-4.3=30.5\), so the 9:00 p.m. temperature is \(30.5\,{}^\circ\text{F}\). b) Highest: \(45.9\,{}^\circ\text{F}\); lowest: \(27.5\,{}^\circ\text{F}\)
5142317
Lucas tracks his allowance account. At the beginning of April, the balance is \(\$28.50\). These transactions occur during the month: <table> <tr><td>Apr. 4</td><td>Movie and popcorn</td><td>\(-\$14.20\)</td></tr> <tr><td>Apr. 10</td><td>Gift from Grandma</td><td>\(+\$20.00\)</td></tr> <tr><td>Apr. 15</td><td>New video game</td><td>\(-\$35.00\)</td></tr> <tr><td>Apr. 22</td><td>Mowing a neighbor's lawn</td><td>\(+\$12.50\)</td></tr> <tr><td>Apr. 28</td><td>Pizza with friends</td><td>\(-\$6.80\)</td></tr> </table> a) Write an expression and find the account balance at the end of April. b) Determine whether the account balance was ever negative during April. If so, give the lowest balance.

Hints

- Treat income as positive and spending as negative. - Apply the transactions in chronological order. - Keep each intermediate balance to check whether the account ever goes below \(0\).

Solution

1. For a), use \(28.50-14.20+20.00-35.00+12.50-6.80\). 2. The balances in order are \(\$28.50\), \(\$14.30\), \(\$34.30\), \(-\$0.70\), \(\$11.80\), and \(\$5.00\). 3. The end-of-month balance is \(\$5.00\). 4. For b), the balance was negative after the Apr. 15 purchase, and the lowest balance was \(-\$0.70\).

Answer

a) \(\$5.00\) b) Yes. The lowest balance was \(-\$0.70\).
5142327
A bulk-food store tracks changes in its oatmeal inventory in pounds. On Monday morning, the store has \(34.0\) pounds. During the week, these amounts are sold (negative) or delivered (positive): <table> <tr><td>Monday</td><td>\(-7.0\,\text{lb}\)</td></tr> <tr><td>Tuesday</td><td>\(-10.0\,\text{lb}\)</td></tr> <tr><td>Wednesday</td><td>\(+44.0\,\text{lb}\) (delivery)</td></tr> <tr><td>Thursday</td><td>\(-16.5\,\text{lb}\)</td></tr> <tr><td>Friday</td><td>\(-20.5\,\text{lb}\)</td></tr> </table> a) How many pounds of oatmeal remain on Friday evening? Write an expression. b) What was the greatest amount of oatmeal in the store at one time during the week?

Hints

- Start with the Monday morning inventory and apply each signed change. - Keep a running list of the inventory after each day. - Compare all the inventory amounts to find the maximum.

Solution

1. For a), \(34.0-7.0-10.0+44.0-16.5-20.5=24.0\). So \(24.0\) pounds remain Friday evening. 2. The successive inventory amounts are \(34.0\), \(27.0\), \(17.0\), \(61.0\), \(44.5\), and \(24.0\) pounds. 3. For b), the greatest inventory amount was \(61.0\) pounds, after Wednesday's delivery.

Answer

a) \(24.0\,\text{lb}\) b) \(61.0\,\text{lb}\)
5178467
In a subtraction expression, the minuend is increased by \(45\). How must the subtrahend change so that the difference is \(60\) greater than it was originally?

Hints

- Find the effect of increasing the minuend by \(45\). - Determine how much additional change is needed to reach \(+60\). - Decide whether increasing or decreasing the subtrahend produces that change.

Solution

1. Increasing the minuend by \(45\) increases the difference by \(45\). 2. The desired total increase is \(60\), so an additional increase of \(60 - 45 = 15\) is needed. 3. Decreasing the subtrahend by \(15\) increases the difference by \(15\): \((m + 45) - (s - 15) = m - s + 60\).

Answer

The subtrahend must decrease by \(15\).
5179107
In a subtraction expression, the subtrahend is \(2458\). It is \(1035\) less than the minuend. a) Find the minuend. b) The minuend stays fixed. By how much must the subtrahend decrease for the difference to increase by \(150\)?

Hints

- Add the difference between the numbers to the smaller number to find the minuend. - When the minuend stays fixed, subtracting less makes the difference greater. - In part b), decide whether you need the original value of the difference.

Solution

1. Since the subtrahend is \(1035\) less than the minuend, the minuend is \(2458 + 1035 = 3493\). 2. With the minuend fixed, decreasing the subtrahend by \(150\) increases the difference by \(150\).

Answer

a) The minuend is \(3493\). b) The subtrahend must decrease by \(150\).
5179817
Three students evaluate \(1000 - 250 - 150 - 100\). - **Leo:** \(1000 - 250 = 750\), \(750 - 150 = 600\), \(600 - 100 = 500\). - **Mia:** \(250 + 150 + 100 = 500\), then \(1000 - 500 = 500\). - **Noah:** \(1000 - 250 = 750\), \(150 - 100 = 50\), then \(750 - 50 = 700\). Decide which methods are correct. Explain Noah’s error if his method is incorrect.

Hints

- Subtraction of several positive amounts can be rewritten by adding the amounts and subtracting their total. - Check whether each rewrite is equivalent to the original expression. - Compare \(-150 - 100\) with \(-(150 - 100)\).

Solution

1. Leo evaluates the subtraction from left to right and obtains \(500\), so his method is correct. 2. Mia combines all three amounts being subtracted: \(1000 - 250 - 150 - 100 = 1000 - (250 + 150 + 100) = 500\). Her method is also correct. 3. Noah replaces subtracting both \(150\) and \(100\) with subtracting their difference. However, \(-150 - 100\) is not equal to \(-(150 - 100)\). His result is incorrect.

Answer

Leo and Mia are correct. Noah is incorrect because he subtracts \(150 - 100\) instead of subtracting both \(150\) and \(100\).
5180037
A magician says, “Start with my secret number. Add \(2450\), subtract \(1100\), and then add \(325\). The result is the least five-digit whole number.” Find the secret number.

Hints

- First identify the least five-digit whole number. - Work backward and use the inverse of each operation. - Keep the place values aligned when calculating.

Solution

1. The least five-digit whole number is \(10{,}000\). 2. Work backward and undo adding \(325\): \(10{,}000 - 325 = 9675\). 3. Undo subtracting \(1100\): \(9675 + 1100 = 10{,}775\). 4. Undo adding \(2450\): \(10{,}775 - 2450 = 8325\). 5. Check: \(8325 + 2450 - 1100 + 325 = 10{,}000\).

Answer

\(8325\)
5180707
A research submersible is at \(-120\,\text{m}\), where sea level is \(0\,\text{m}\). It descends another \(150\,\text{m}\) to take a measurement, then rises \(40\,\text{m}\). What is the submersible's final position relative to sea level?

Hints

- Represent sea level with \(0\) on a number line. - How do descending and rising affect a signed number? - Break the motion into two calculations.

Solution

1. After descending, the position is \(-120-150=-270\), or \(-270\,\text{m}\). 2. After rising, the position is \(-270+40=-230\), or \(-230\,\text{m}\).

Answer

The submersible is at \(-230\,\text{m}\), which is \(230\,\text{m}\) below sea level.
5181037
Choose signs for \(20\) and \(45\) to make each equation true. Write the completed equation. a) \((\mathbin{\square}20)+(\mathbin{\square}45)=65\) b) \((\mathbin{\square}20)+(\mathbin{\square}45)=-25\) c) \((\mathbin{\square}20)+(\mathbin{\square}45)=25\) d) \((\mathbin{\square}20)+(\mathbin{\square}45)=-65\)

Hints

- Decide whether the target uses the sum or difference of the absolute values. - Equal signs add absolute values; unlike signs subtract them. - The addend with the greater absolute value determines the sign of a nonzero sum.

Solution

1. To get \(65\), both addends are positive: \(20+45=65\). 2. To get \(-25\), the greater absolute value must be negative: \(20+(-45)=-25\). 3. To get \(25\), the greater absolute value must be positive: \(-20+45=25\). 4. To get \(-65\), both addends are negative: \(-20+(-45)=-65\).

Answer

a) \(20+45=65\) b) \(20+(-45)=-25\) c) \(-20+45=25\) d) \(-20+(-45)=-65\)
5181267
Insert either \(+\) or \(-\) in each box to make the equation true. a) \((-18)\mathbin{\square}(-7)=-11\) b) \((+25)\mathbin{\square}(-5)=+30\) c) \((-40)\mathbin{\square}(+10)=-50\) d) \((+12)\mathbin{\square}(-12)=0\)

Hints

- Distinguish the operation sign between the numbers from the sign attached to each number. - Subtracting a negative number is the same as adding its opposite. - Test both operations and check the result.

Solution

1. In a), \(-18-(-7)=-11\), so the operation is subtraction. 2. In b), \(25-(-5)=30\), so the operation is subtraction. 3. In c), \(-40-(+10)=-50\), so the operation is subtraction. 4. In d), \(12+(-12)=0\), so the operation is addition.

Answer

a) \(-\) b) \(-\) c) \(-\) d) \(+\)
5181277
Insert the missing signs so that each equation is true. a) \((\bigcirc100)+(\bigcirc40)+(-20)=-80\) b) \((+50)+(\bigcirc30)+(\bigcirc10)=+10\)

Hints

- Work through each sum from left to right. - Compare the starting total with the target total. - Combine the terms whose signs are already known before choosing the missing signs.

Solution

1. In a), \(-100+40-20=-80\), so the signs are \(-\) and \(+\). 2. In b), \(50-30-10=10\), so both signs are \(-\).

Answer

a) \((-100)+(+40)+(-20)=-80\) b) \((+50)+(-30)+(-10)=+10\)
5181417
Estimate to decide which expression has the lesser value. Then calculate both values exactly. Expression 1: \(-3250+(-1780)\) Expression 2: \(-8410+3520\)

Hints

- Round each addend to the nearest thousand for the estimate. - Evaluate the expressions separately. - The negative value farther left is less.

Solution

1. Rounding to the nearest thousand gives about \(-3000+(-2000)=-5000\) for Expression 1 and \(-8000+4000=-4000\) for Expression 2, so Expression 1 should be less. 2. Exactly, Expression 1 is \(-3250-1780=-5030\). 3. Expression 2 is \(-8410+3520=-4890\). 4. Since \(-5030<-4890\), Expression 1 is less.

Answer

Expression 1 is less. Its exact value is \(-5030\), and Expression 2 equals \(-4890\).
5181437
Estimate by rounding each addend to the nearest thousand, and then evaluate \(-4890+(-3120)+6950\) exactly.

Hints

- Round each number to the nearest thousand. - Add the two negative numbers first. - Check that the exact value is close to the estimate.

Solution

1. The estimate is \(-5000+(-3000)+7000=-1000\). 2. The exact sum of the first two addends is \(-4890+(-3120)=-8010\). 3. Then \(-8010+6950=-1060\).

Answer

Estimate: \(-1000\) Exact value: \(-1060\)
5181587
Write an expression for each description and evaluate it. a) Find the difference of \(1500\) and \(850\), then subtract the opposite of \(150\). b) Add the absolute value of \(-75\) to the sum of \(-125\) and \(-200\).

Hints

- Evaluate the difference, sum, absolute value, and opposite separately. - Subtracting a negative is adding a positive. - Follow the order stated in each description.

Solution

1. In part a), \(1500-850=650\), and the opposite of \(150\) is \(-150\). Thus, \(650-(-150)=800\). 2. In part b), \(-125+(-200)=-325\), and \(|-75|=75\). Thus, \(-325+75=-250\).

Answer

a) \((1500-850)-(-150)=800\) b) \([-125+(-200)]+|-75|=-250\)
5181597
Subtract the sum of \(-48\) and \(122\) from the difference of \(300\) and \(-50\). Write and evaluate the expression.

Hints

- Identify which quantity is being subtracted from which. - Evaluate the two grouped expressions separately. - Then subtract the second result from the first.

Solution

1. The difference is \(300-(-50)=350\). 2. The sum is \(-48+122=74\). 3. Subtracting the sum from the difference gives \(350-74=276\).

Answer

\([300-(-50)]-[-48+122]=276\)
5181677
Match each sum to its exact result. Use estimation to find the pairs efficiently. Sums: a) \(-298+(-405)+(-102)\) b) \(-895+302+198\) c) \(512+(-251)+(-262)\) d) \(-1203+(-395)+(-102)\) Results: 1) \(-1\) 2) \(-1700\) 3) \(-805\) 4) \(-395\)

Hints

- Round to nearby hundreds. - Track the signs carefully. - Match each estimate before checking exactly.

Solution

1. Part a) is about \(-300-400-100=-800\), and exactly equals \(-805\), so it matches 3. 2. Part b) is about \(-900+300+200=-400\), and exactly equals \(-395\), so it matches 4. 3. Part c) is about \(500-250-250=0\), and exactly equals \(-1\), so it matches 1. 4. Part d) is about \(-1200-400-100=-1700\), and exactly equals \(-1700\), so it matches 2.

Answer

a) 3 b) 4 c) 1 d) 2
5181797
Decide whether each statement is true or false. Briefly explain. a) Start with any number, add \(12\), and then add the opposite of the original number. The result is always \(12\). b) The sum of a negative number and its opposite is always positive.

Hints

- Represent the original number with a variable. - Group a number with its opposite. - Remember that zero is not positive.

Solution

1. For any number \(n\), \(n+12+(-n)=12+[n+(-n)]=12\), so a) is true. 2. A number and its opposite sum to \(0\), which is neither positive nor negative. Therefore, b) is false.

Answer

a) True; \(n+12+(-n)=12\). b) False; a number and its opposite sum to \(0\).
5181897
Maya claims, “The sum of two negative integers is always less than either addend.” Test the claim using \(-7+(-3)\), and then explain whether it is always true.

Hints

- Calculate the example first. - Compare the sum with both addends. - Think about the direction of movement caused by adding a negative number.

Solution

1. \(-7+(-3)=-10\), and \(-10<-7\) and \(-10<-3\). 2. In general, adding a negative number moves left on a number line. Therefore, the sum of two negative integers lies to the left of each addend and is less than both.

Answer

The example gives \(-10\), which is less than both \(-7\) and \(-3\). The claim is always true because adding a negative integer decreases a number.
5181907
Let \(a=-8\) in \(a+b=s\). Give one integer value of \(b\) for each condition. a) \(s<a\) b) \(s=a\) c) \(s>a\)

Hints

- A negative addend moves the sum left. - Zero leaves a value unchanged. - A positive addend moves the sum right.

Solution

1. To make the sum less than \(-8\), choose any negative integer, such as \(b=-2\), giving \(-10<-8\). 2. To leave the value unchanged, choose \(b=0\). 3. To make the sum greater than \(-8\), choose any positive integer, such as \(b=5\), giving \(-3>-8\).

Answer

a) One answer is \(b=-2\). b) \(b=0\) c) One answer is \(b=5\).
5181937
Give one pair of integers with sum \(-32\) for each condition. a) Both addends are negative. b) The first addend is positive, and the second is negative. c) The first addend is negative, and the second is positive.

Hints

- For two negative addends, their absolute values must total \(32\). - With unlike signs, the negative addend must have the greater absolute value. - Many answers are possible.

Solution

1. Two negative addends can be \(-16\) and \(-16\), since \(-16+(-16)=-32\). 2. A positive first addend and negative second addend can be \(8\) and \(-40\), since \(8+(-40)=-32\). 3. A negative first addend and positive second addend can be \(-50\) and \(18\), since \(-50+18=-32\).

Answer

a) \(-16+(-16)=-32\) b) \(8+(-40)=-32\) c) \(-50+18=-32\)
5181947
A submarine's net vertical change is \(-80\,\text{m}\), where positive changes mean upward movement and negative changes mean downward movement. Give one addition equation for each situation. a) The submarine descends during both stages. b) The submarine rises during the first stage and descends during the second stage.

Hints

- Represent downward movement with a negative change and upward movement with a positive change. - Choose the first change, then find the second change needed for a total of \(-80\,\text{m}\). - Many answers are possible.

Solution

1. Two downward changes can be \(-30\,\text{m}\) and \(-50\,\text{m}\), giving \(-30+(-50)=-80\). 2. An upward change followed by a larger downward change can be \(20\,\text{m}\) and \(-100\,\text{m}\), giving \(20+(-100)=-80\).

Answer

a) One example is \((-30\,\text{m})+(-50\,\text{m})=-80\,\text{m}\). b) One example is \(20\,\text{m}+(-100\,\text{m})=-80\,\text{m}\).
5181957
Complete each equation so the sum is \(-50\). a) \(25+\square=-50\) b) \(-70+\square=-50\) c) \(\square+\square=-50\), using two identical addends.

Hints

- Use subtraction to find a missing addend. - Check the direction from the known value to \(-50\). - For equal addends, divide the target sum by \(2\).

Solution

1. In part a), the missing addend is \(-50-25=-75\). 2. In part b), the missing addend is \(-50-(-70)=20\). 3. Two equal addends must each be half of \(-50\), so each is \(-25\).

Answer

a) \(25+(-75)=-50\) b) \(-70+20=-50\) c) \(-25+(-25)=-50\)
5181987
Evaluate this claim: “Subtracting the absolute value of an integer from the integer always gives \(0\).” Decide whether the claim is true or false and justify your answer with examples.

Hints

- Test both a positive and a negative integer. - Remember that absolute value is always nonnegative. - An “always” claim is false if one counterexample exists.

Solution

1. For a positive integer such as \(7\), \(7-|7|=7-7=0\). 2. For a negative integer such as \(-4\), \(-4-|-4|=-4-4=-8\), not \(0\). 3. Because one counterexample is enough to disprove an “always” claim, the claim is false.

Answer

The claim is false. For example, \(-5-|-5|=-5-5=-10\), which is not \(0\).
5182407
Complete the calculation chain \(-120\xrightarrow{+45}\square\xrightarrow{-30}\square\xrightarrow{-(-60)}\square\).

Hints

- Work from left to right. - Treat each arrow as a new operation. - Subtracting a negative is adding a positive.

Solution

1. \(-120+45=-75\). 2. \(-75-30=-105\). 3. \(-105-(-60)=-105+60=-45\).

Answer

The intermediate values are \(-75\) and \(-105\), and the final value is \(-45\).
5182487
Each equation is missing both a sign attached to a number and an operation sign. Insert \(+\) or \(-\) to make each equation true. Find all possibilities. a) \((\square8)\mathbin{\bigcirc}(-12)=+20\) b) \((-25)\mathbin{\bigcirc}(\square15)=-10\) c) \((\square40)\mathbin{\bigcirc}(+60)=-20\)

Hints

- Test the possible sign and operation combinations systematically. - Keep the sign attached to the number separate from the operation sign. - Verify every combination that works so you find all solutions.

Solution

1. In a), the only solution is \((+8)-(-12)=20\). 2. In b), both \((-25)+(+15)=-10\) and \((-25)-(-15)=-10\) work. 3. In c), the only solution is \((+40)-(+60)=-20\).

Answer

a) \((+8)-(-12)=+20\) b) \((-25)+(+15)=-10\) or \((-25)-(-15)=-10\) c) \((+40)-(+60)=-20\)
5182547
Complete the subtraction table. Each entry is the row value minus the column value. <table> <tr> <th colspan="2" rowspan="2"></th> <th colspan="4">Number being subtracted</th> </tr> <tr> <th>\(15\)</th> <th>\(-10\)</th> <th>\(-35\)</th> <th>\(60\)</th> </tr> <tr> <th rowspan="3">Starting number</th> <th>\(-20\)</th> <td></td> <td></td> <td></td> <td></td> </tr> <tr> <th>\(45\)</th> <td></td> <td></td> <td></td> <td></td> </tr> <tr> <th>\(-80\)</th> <td></td> <td></td> <td></td> <td></td> </tr> </table>

Hints

- Subtracting a negative number is the same as adding its opposite. - Think about which direction subtraction moves on a number line. - A result can be negative when the number being subtracted is greater than the starting number. - Work through one row at a time.

Solution

1. For the row beginning with \(-20\): \(-20-15=-35\), \(-20-(-10)=-10\), \(-20-(-35)=15\), and \(-20-60=-80\). 2. For the row beginning with \(45\): \(45-15=30\), \(45-(-10)=55\), \(45-(-35)=80\), and \(45-60=-15\). 3. For the row beginning with \(-80\): \(-80-15=-95\), \(-80-(-10)=-70\), \(-80-(-35)=-45\), and \(-80-60=-140\).

Answer

The completed table is: <table> <tr> <th colspan="2"></th> <th>\(15\)</th> <th>\(-10\)</th> <th>\(-35\)</th> <th>\(60\)</th> </tr> <tr> <th rowspan="3"></th> <th>\(-20\)</th> <td>\(-35\)</td> <td>\(-10\)</td> <td>\(15\)</td> <td>\(-80\)</td> </tr> <tr> <th>\(45\)</th> <td>\(30\)</td> <td>\(55\)</td> <td>\(80\)</td> <td>\(-15\)</td> </tr> <tr> <th>\(-80\)</th> <td>\(-95\)</td> <td>\(-70\)</td> <td>\(-45\)</td> <td>\(-140\)</td> </tr> </table>
5182907
Evaluate \((-128)-[(+52)+(-97)]\).

Hints

- Use the order of operations. - Evaluate the expression inside the brackets first. - Replace the bracketed expression with its value before continuing.

Solution

1. Evaluate inside the brackets: \(52+(-97)=-45\). 2. Substitute the result: \(-128-(-45)\). 3. Rewrite the subtraction and evaluate: \(-128+45=-83\).

Answer

\(-83\)
5182917
Evaluate \([(-34)+(+86)]-[(-42)-(+58)]\) step by step.

Hints

- Break the expression into two smaller calculations. - Evaluate each bracketed expression separately. - Keep track of both intermediate results before the final subtraction. - Subtracting a negative number is the same as adding a positive number.

Solution

1. Evaluate the first bracketed expression: \(-34+86=52\). 2. Evaluate the second bracketed expression: \(-42-58=-100\). 3. Subtract the two results: \(52-(-100)=52+100=152\).

Answer

\(152\)
5182967
Use the numbers \(-34\) and \(-16\). a) Find their sum. b) Subtract the sum from the opposite of \(50\). What is the result?

Hints

- A sum is the result of addition. - The opposite of a number is the same distance from zero on the other side. - Pay close attention to the signs and parentheses in part b).

Solution

1. The sum is \((-34)+(-16)=-50\). 2. The opposite of \(50\) is \(-50\). 3. Subtract the sum from that number: \(-50-(-50)=0\).

Answer

a) \(-50\) b) \(0\)
5182977
Use the numbers \(12\) and \(-18\). First find their sum and their difference, in that order. Then subtract the difference from the sum. What is the final result?

Hints

- Find the sum and the difference separately. - Subtracting a negative number is the same as adding a positive number. - Read carefully to determine which value is subtracted from which.

Solution

1. The sum is \(12+(-18)=-6\). 2. The difference is \(12-(-18)=30\). 3. Subtract the difference from the sum: \(-6-30=-36\).

Answer

\(-36\)
5183027
On a cold January day, temperatures are recorded at two locations in the mountains. At noon, the mountain station is \(-11\,^\circ\text{F}\), and at night it is \(-24\,^\circ\text{F}\). In the valley, the temperature is \(4\,^\circ\text{F}\) at noon and \(-6\,^\circ\text{F}\) at night. a) For each location, by how many degrees Fahrenheit did the temperature decrease from noon to night? b) Find the temperature difference between the mountain station and the valley at noon and at night.

Hints

- Picture the temperatures on a vertical number line, like a thermometer. - A temperature difference is the distance between two values. - Notice whether both values are negative or whether the interval crosses zero. - To find the size of a decrease, subtract the lower final value from the higher initial value.

Solution

1. At the mountain station, the decrease is \(-11-(-24)=13\), so the temperature decreased by \(13\,^\circ\text{F}\). 2. In the valley, the decrease is \(4-(-6)=10\), so the temperature decreased by \(10\,^\circ\text{F}\). 3. At noon, the difference is \(4-(-11)=15\,^\circ\text{F}\). 4. At night, the difference is \(-6-(-24)=18\,^\circ\text{F}\).

Answer

a) Mountain station: \(13\,^\circ\text{F}\); valley: \(10\,^\circ\text{F}\) b) Noon: \(15\,^\circ\text{F}\); night: \(18\,^\circ\text{F}\)
5183047
A laboratory uses several cooling areas. The room temperature is \(22\,^\circ\text{C}\), a refrigerator is \(4\,^\circ\text{C}\), a freezer is \(-18\,^\circ\text{C}\), and a dry-ice chamber is \(-78\,^\circ\text{C}\). a) Write and evaluate an absolute-difference expression for the room-to-freezer temperature difference. b) Write and evaluate an absolute-difference expression for the refrigerator-to-dry-ice-chamber difference. c) Write absolute-difference expressions for the refrigerator-to-freezer and freezer-to-dry-ice-chamber differences. Which is greater?

Hints

- Use the same absolute-difference structure for every temperature separation. - Keep the signed temperatures inside the expression before taking absolute value. - Compare the resulting nonnegative distances in part c).

Solution

1. a) \(|22-(-18)|=40\,^\circ\text{C}\). 2. b) \(|4-(-78)|=82\,^\circ\text{C}\). 3. c) \(|4-(-18)|=22\,^\circ\text{C}\) and \(|-18-(-78)|=60\,^\circ\text{C}\). The second difference is greater.

Answer

a) \(|22-(-18)|=40\,^\circ\text{C}\) b) \(|4-(-78)|=82\,^\circ\text{C}\) c) \(|4-(-18)|=22\,^\circ\text{C}\) and \(|-18-(-78)|=60\,^\circ\text{C}\); the freezer-to-dry-ice-chamber difference is greater.
5183107
Complete the account table. In each row, an amount is withdrawn from the previous balance to produce the new balance. <table> <tr><th>Previous balance</th><th>Withdrawal</th><th>New balance</th></tr> <tr><td>\(\$12\)</td><td>\(\$20\)</td><td>(1)</td></tr> <tr><td>\(-\$8\)</td><td>\(\$15\)</td><td>(2)</td></tr> <tr><td>(3)</td><td>\(\$25\)</td><td>\(-\$10\)</td></tr> </table>

Hints

- A withdrawal decreases the balance. - Use an inverse operation to find a missing previous balance. - Pay close attention to the sign when the account already has a negative balance.

Solution

1. For (1), \(12-20=-8\), so the new balance is \(-\$8\). 2. For (2), \(-8-15=-23\), so the new balance is \(-\$23\). 3. For (3), let the previous balance be \(x\). Solve \(x-25=-10\), which gives \(x=15\).

Answer

(1) \(-\$8\) (2) \(-\$23\) (3) \(\$15\)
5183217
Evaluate each difference, and then insert \(<\), \(>\), or \(=\) to make a true statement. a) \((-12)-8\mathbin{\square}-12\) b) \((-12)-(-8)\mathbin{\square}-12\) c) \(15-15\mathbin{\square}15\) d) \(15-(-15)\mathbin{\square}15\)

Hints

- Evaluate the expression on the left first. - Pay close attention when subtracting a negative number. - On a number line, the number farther right is greater.

Solution

1. In a), \(-12-8=-20\), and \(-20<-12\). 2. In b), \(-12-(-8)=-4\), and \(-4>-12\). 3. In c), \(15-15=0\), and \(0<15\). 4. In d), \(15-(-15)=30\), and \(30>15\).

Answer

a) \(<\) b) \(>\) c) \(<\) d) \(>\)
5183227
Consider the two expressions. Expression \(A\): \((-40)-25\) Expression \(B\): \((-40)-(-25)\) Without calculating exact values, explain which expression is greater. Use what you know about subtracting integers and compare each result with \(-40\).

Hints

- Think about movement on a number line rather than calculating. - What happens when a positive number is subtracted? - What happens when a negative number is subtracted? - Compare both changes from the same starting value, \(-40\).

Solution

1. In Expression \(A\), a positive number is subtracted from \(-40\), so the result is less than \(-40\). 2. In Expression \(B\), a negative number is subtracted. This is equivalent to adding \(25\), so the result is greater than \(-40\). 3. Therefore, Expression \(B\) is greater.

Answer

Expression \(B\) is greater. Subtracting \(25\) makes \(-40\) smaller, while subtracting \(-25\) is the same as adding \(25\), which makes \(-40\) greater.
5183667
Evaluate each expression with three numbers. First rewrite it without parentheses using simplified notation. a) \((-15)+(+25)-(-10)\) b) \((+40)-(+60)+(-20)\) c) \((-100)-(-30)-(+70)\)

Hints

- Rewrite all signed numbers before calculating. - Subtracting a negative number becomes addition. - After simplifying the notation, combine the numbers from left to right or in convenient groups.

Solution

1. In a), \(-15+25+10=20\). 2. In b), \(40-60-20=-40\). 3. In c), \(-100+30-70=-140\).

Answer

a) \(-15+25+10=20\) b) \(40-60-20=-40\) c) \(-100+30-70=-140\)
5183677
Start at \(-50\) and perform these operations in order: 1. Subtract \(-30\). 2. Add \(-45\). 3. Subtract \(+15\). Write one expression with parentheses, simplify it, and find the final value.

Hints

- Translate each instruction into one operation in a single expression. - Subtracting a negative number becomes addition. - Perform the operations in order after simplifying the signs.

Solution

1. The expression is \(-50-(-30)+(-45)-(+15)\). 2. Simplify the signs: \(-50+30-45-15\). 3. Evaluate: \(-50+30=-20\), \(-20-45=-65\), and \(-65-15=-80\).

Answer

Expression: \(-50-(-30)+(-45)-(+15)\) Simplified: \(-50+30-45-15\) Final value: \(-80\)
5183837
Rewrite the expression as a sum. Then show an equivalent rearrangement and grouping that creates convenient partial sums before finding the final value. \(-2650+1433-350+567\)

Hints

- Rewrite every subtraction as addition of the opposite. - Keep each sign attached to its term while rearranging. - Compare possible pairings and choose ones that make the arithmetic simpler.

Solution

1. Rewrite subtraction as addition of the opposite: \(-2650+1433+(-350)+567\). 2. Rearrange and group: \((-2650-350)+(1433+567)\). 3. Evaluate the groups: \(-3000+2000=-1000\).

Answer

Sum form: \(-2650+1433+(-350)+567\) One valid efficient regrouping: \((-2650-350)+(1433+567)\) Value: \(-1000\)
5183857
Compare the values and insert \(<\), \(>\), or \(=\) in each box. a) \(18-30\mathbin{\square}-18+30\) b) \(-50-(-20)\mathbin{\square}-50+20\) c) \(45-(10+50)\mathbin{\square}45-10-50\) d) \(-(25-5)\mathbin{\square}-25-5\)

Hints

- Evaluate the left and right sides separately. - Pay close attention to a negative sign outside parentheses. - On a number line, the number farther right is greater.

Solution

1. In a), the values are \(-12\) and \(12\), so use \(<\). 2. In b), both sides equal \(-30\), so use \(=\). 3. In c), both sides equal \(-15\), so use \(=\). 4. In d), the values are \(-20\) and \(-30\), so use \(>\).

Answer

a) \(<\) b) \(=\) c) \(=\) d) \(>\)
5183867
Three expressions have the same value. Which expression does not belong? Justify your choice by evaluating all four. A) \(-22+10-8\) B) \(-22+(8-10)\) C) \(10-(22+8)\) D) \(-(22-10)-8\)

Hints

- Evaluate each expression carefully using the order of operations. - Record the sign of each intermediate value. - Compare all four final values.

Solution

1. A has value \(-22+10-8=-20\). 2. B has value \(-22+(8-10)=-24\). 3. C has value \(10-(22+8)=-20\). 4. D has value \(-(22-10)-8=-20\). 5. Expression B is the only expression with a different value.

Answer

Expression B does not belong. Its value is \(-24\), while A, C, and D each have value \(-20\).
5183917
Rewrite each expression as a sum, and then evaluate. a) \(25{,}000-60{,}000\) b) \(-10^4-5000\) c) \(8800-(-1200)\) d) \(-10^6-(-10^5)\)

Hints

- Write each power of ten as an ordinary number first. - Keep track of the zeros in large numbers. - To subtract a number, add its opposite.

Solution

1. In a), \(25{,}000+(-60{,}000)=-35{,}000\). 2. In b), \(-10^4=-10{,}000\), so \((-10{,}000)+(-5000)=-15{,}000\). 3. In c), \(8800+1200=10{,}000\). 4. In d), \(-10^6=-1{,}000{,}000\) and \(-(-10^5)=100{,}000\), so \(-1{,}000{,}000+100{,}000=-900{,}000\).

Answer

a) \(-35{,}000\) b) \(-15{,}000\) c) \(10{,}000\) d) \(-900{,}000\)
5184017
Mr. Schmidt pays a \(\$545\) bill from his checking account. Soon afterward, a credit of \(\$210\) is deposited. His balance is then exactly \(-\$185\). What was his account balance before these two transactions?

Hints

- Determine the combined effect of the two transactions. - You can also undo the transactions in reverse order, starting from the final balance. - A bill decreases the balance, while a credit increases it.

Solution

1. The net change is \(-545+210=-335\). 2. Let \(x\) be the initial balance. Then \(x-335=-185\). 3. Add \(335\) to both sides: \(x=-185+335=150\). 4. Check: \(150-545+210=-185\).

Answer

The account balance before the transactions was \(\$150\).
5184027
An elevator in a parking garage first travels up \(5\) levels and then down \(8\) levels. It ends on Level \(-2\). On which level did the elevator start?

Hints

- Picture the garage levels as a vertical number line. - Combine the upward and downward movements into one net change. - Starting from the ending level, reverse the net change.

Solution

1. The total change is \(5-8=-3\), so the elevator ends \(3\) levels below where it started. 2. Let \(x\) be the starting level. Then \(x-3=-2\). 3. Add \(3\) to both sides: \(x=1\).

Answer

The elevator started on Level \(1\).
5184057
First estimate by rounding each number to the nearest ten. Then evaluate efficiently by rearranging and grouping terms. \(48+63-28+37-80\)

Hints

- Look for numbers that combine to make convenient multiples of \(10\) or \(100\). - When rearranging, keep each number together with the sign before it. - Numbers with the same ones digit may be easy to subtract.

Solution

1. Estimate: \(50+60-30+40-80=40\). 2. Rearrange and group: \((48-28)+(63+37)-80\). 3. Evaluate: \(20+100-80=40\).

Answer

Estimate: \(40\) Exact value: \(40\)
5184067
First make a reasonable estimate. Then evaluate efficiently by grouping positive and negative numbers. \(-135+420-65+580-300\)

Hints

- Group the terms with negative signs. - Look for pairs that combine to make convenient multiples of \(100\). - Then combine the positive total and the negative total.

Solution

1. Estimate by rounding to the nearest hundred: \(-100+400-100+600-300=500\). 2. Group the negative terms: \(-135-65=-200\). 3. Group the positive terms: \(420+580=1000\). 4. Combine the results: \(1000-200-300=500\).

Answer

Estimate: \(\approx 500\) Exact value: \(500\)
5184077
Make a reasonable estimate, and then evaluate efficiently by rearranging or grouping terms. \(2750-438-1250+638-500\)

Hints

- Look for numbers with matching ending digits. - Keep each number together with the sign before it when rearranging. - Group terms so that the intermediate values are easy to calculate.

Solution

1. Estimate by rounding to the nearest hundred: \(2800-400-1300+600-500=1200\). 2. Rearrange and group: \((2750-1250)+(638-438)-500\). 3. Evaluate: \(1500+200-500=1200\).

Answer

Estimate: \(\approx 1200\) Exact value: \(1200\)
5184107
Rewrite without parentheses using simplified notation, and then evaluate: \(1000-(-450)+(-1250)-(+200)\).

Hints

- Replace each subtraction of a negative number with addition. - Group positive and negative terms separately. - Equal opposite amounts cancel to zero.

Solution

1. Rewrite the expression as \(1000+450-1250-200\). 2. Group the positive terms and the negative terms: \((1000+450)-(1250+200)\). 3. Both groups equal \(1450\), so the result is \(0\).

Answer

\(1000+450-1250-200=0\)
5184227
Consider the expression \(-45+128-55+72\). a) Evaluate the expression efficiently by grouping terms. b) Replace the last number so that the expression has a value of \(0\). What should the new number be?

Hints

- Look for pairs that combine easily. - Keep track of the sign before each number. - To make the total zero, the last number must be the opposite of the sum of the first three terms.

Solution

1. For a), group convenient terms: \((-45-55)+(128+72)=-100+200=100\). 2. For b), the first three terms total \(-45+128-55=28\). 3. The new last number \(x\) must satisfy \(28+x=0\), so \(x=-28\).

Answer

a) \(100\) b) \(-28\)
5184237
Consider the expression \(1250-3400+750-600\). a) Evaluate it efficiently by grouping terms. b) Replace the last term, \(-600\), so that the expression has a value of \(0\). Give the new signed number.

Hints

- Add the positive terms first. - Find the value of the expression without the final term. - To make the total zero, use the opposite of that value.

Solution

1. For a), group the positive terms: \((1250+750)-3400-600=2000-4000=-2000\). 2. For b), the first three terms total \(1250-3400+750=-1400\). 3. The new last term must be \(1400\), because \(-1400+1400=0\).

Answer

a) \(-2000\) b) \(+1400\)
5184277
Evaluate each expression step by step. Follow the order of operations. a) \(26-[(41-55)+(82-14)]\) b) \([-18+(72-95)]-(31+29)\)

Hints

- Evaluate the innermost parentheses first. - Replace each set of parentheses with its value before continuing. - Keep careful track of signs in the final subtraction.

Solution

1. In a), \(41-55=-14\) and \(82-14=68\). Then \(-14+68=54\), so \(26-54=-28\). 2. In b), \(72-95=-23\). Then \(-18+(-23)=-41\), and \(31+29=60\). Finally, \(-41-60=-101\).

Answer

a) \(-28\) b) \(-101\)
5184287
Evaluate each expression using the order of operations. a) \(-32-(74+26)+(-85+35)\) b) \((-94-126)-(-52+318)\)

Hints

- Evaluate each parenthetical expression first. - Pay close attention to subtraction between two grouped expressions. - Write down intermediate values to reduce sign errors.

Solution

1. In a), \(74+26=100\) and \(-85+35=-50\). Then \(-32-100+(-50)=-182\). 2. In b), \(-94-126=-220\) and \(-52+318=266\). Then \(-220-266=-486\).

Answer

a) \(-182\) b) \(-486\)
5184317
Evaluate \(-240+115+(-60)+85\) efficiently. Name the properties used to regroup the addends.

Hints

- Look for pairs that make convenient hundreds. - Addition allows addends to be reordered and regrouped. - Combine the negative and positive partial sums last.

Solution

1. Use the commutative property to reorder and the associative property to regroup: \([-240+(-60)]+[115+85]\). 2. The partial sums are \(-300\) and \(200\). 3. Therefore, the value is \(-300+200=-100\).

Answer

The value is \(-100\). The commutative and associative properties of addition are used.
5184327
Evaluate efficiently by rearranging and grouping terms: \(781-495-281+195-100\).

Hints

- Look for numbers with matching ending digits. - Keep each number together with its sign when rearranging. - Form groups that are easy to evaluate mentally.

Solution

1. Rewrite the expression as a sum: \(781+(-495)+(-281)+195+(-100)\). 2. Rearrange and group: \((781-281)+(-495+195)-100\). 3. Evaluate: \(500-300-100=100\).

Answer

\(100\)
5184367
Evaluate \((12-45)-(150+5)\).

Hints

- Evaluate the two parenthetical expressions separately. - In the final step, a positive number is being subtracted from a negative number.

Solution

1. Evaluate the first parentheses: \(12-45=-33\). 2. Evaluate the second parentheses: \(150+5=155\). 3. Subtract the results: \(-33-155=-188\).

Answer

\(-188\)
5184607
During one level of a video game, Tim's score changes several times. He first earns \(120\) bonus points, then loses \(250\) points, and finally earns another \(40\) points. His score at the end of the level is \(-30\). What was Tim's score at the beginning of the level?

Hints

- Calculate the total change in the score. - You can work backward from the final score one change at a time. - Decide whether the beginning score must have been greater or less than the ending score.

Solution

1. Find the total change in the score: \(120-250+40=-90\). 2. Let \(x\) be the beginning score. Then \(x-90=-30\). 3. Add \(90\) to both sides: \(x=-30+90=60\).

Answer

Tim began the level with \(60\) points.
5185017
At \(6{:}00\) a.m., the temperature is \(-4\,^\circ\text{F}\). It then rises \(7\,^\circ\text{F}\), falls \(2\,^\circ\text{F}\), rises \(3\,^\circ\text{F}\), falls \(8\,^\circ\text{F}\), and falls another \(5\,^\circ\text{F}\). Find the final temperature. Calculate efficiently.

Hints

- Group positive and negative changes. - Look for values that cancel. - Combine the remaining changes with the starting temperature.

Solution

1. The total is \(-4+7-2+3-8-5\). 2. Group \(7+3=10\) and \(-2-8=-10\), which cancel. 3. The remaining value is \(-4-5=-9\).

Answer

\(-9\,^\circ\text{F}\)
5185027
A research submarine starts at \(-1250\,\text{m}\) relative to sea level. It rises \(450\,\text{m}\), descends \(200\,\text{m}\), rises \(800\,\text{m}\), and descends \(150\,\text{m}\). Find its new position. Calculate efficiently.

Hints

- Combine the upward changes. - Compare their total with the starting depth. - Then account for the remaining descents.

Solution

1. The total is \(-1250+450-200+800-150\). 2. The rises total \(450+800=1250\), which cancels the starting position. 3. The remaining changes are \(-200-150=-350\).

Answer

The submarine is at \(-350\,\text{m}\), or \(350\,\text{m}\) below sea level.
5185037
A club account starts at \(-\$240\). During the quarter, it receives \(\$1550\) in dues, pays \(\$600\) in rent, spends \(\$760\) on equipment, receives a \(\$450\) donation, and pays \(\$300\) for an event. Find the new balance. Calculate efficiently.

Hints

- Group income and expenses. - Look for amounts that make convenient hundreds or thousands. - Combine all signed changes with the starting balance.

Solution

1. The balance is \(-240+1550-600-760+450-300\). 2. Group \(-240-760=-1000\), \(1550+450=2000\), and \(-600-300=-900\). 3. Then \(2000-1000-900=100\).

Answer

The new balance is \(\$100\).
5185057
Let \(a=-45\), \(b=20\), and \(c=-30\). a) Write and evaluate an expression that subtracts \(c\) from the sum of \(a\) and \(b\). b) Write and evaluate an expression that subtracts the sum of \(a\) and \(b\) from \(c\). c) Compare the results. What general observation can you make?

Hints

- Write each expression with variables before substituting values. - Use parentheses around the sum that is subtracted as a group. - Compare the two results and their signs.

Solution

1. In a), \((a+b)-c=(-45+20)-(-30)=-25+30=5\). 2. In b), \(c-(a+b)=-30-(-45+20)=-30-(-25)=-5\). 3. The results are opposites. In general, reversing the order of a subtraction changes the sign of the difference.

Answer

a) \((-45+20)-(-30)=5\) b) \(-30-(-45+20)=-5\) c) The results are opposites. Reversing the order of a subtraction negates the difference.
5185597
A research team measures elevations relative to sea level. <table> <tr><td>Location</td><td>Elevation</td></tr> <tr><td>A (peak)</td><td>\(+340\,\text{m}\)</td></tr> <tr><td>B (meadow)</td><td>\(+115\,\text{m}\)</td></tr> <tr><td>C (village)</td><td>\(-25\,\text{m}\)</td></tr> <tr><td>D (lakeshore)</td><td>\(-142\,\text{m}\)</td></tr> <tr><td>E (cave floor)</td><td>\(-205\,\text{m}\)</td></tr> </table> For each part, write and evaluate an absolute-difference expression. a) Find the elevation difference between A and C. b) Find the vertical separation between D and E. c) Find the elevation difference between the highest and lowest locations.

Hints

- Use absolute difference even when you already know which elevation is higher. - Keep negative elevations signed inside the subtraction. - For part c), identify the extreme elevations before writing the expression.

Solution

1. a) \(|340-(-25)|=365\,\text{m}\). 2. b) \(|-142-(-205)|=63\,\text{m}\). 3. c) The highest is A and lowest is E, so \(|340-(-205)|=545\,\text{m}\).

Answer

a) \(|340-(-25)|=365\,\text{m}\) b) \(|-142-(-205)|=63\,\text{m}\) c) \(|340-(-205)|=545\,\text{m}\)
5185607
Ms. Berger records her checking account balance at the end of each weekday. <table> <tr><td>Monday</td><td>\(+\$12\)</td></tr> <tr><td>Tuesday</td><td>\(-\$4\)</td></tr> <tr><td>Wednesday</td><td>\(-\$18\)</td></tr> <tr><td>Thursday</td><td>\(-\$9\)</td></tr> <tr><td>Friday</td><td>\(+\$22\)</td></tr> </table> a) Find the signed change from Monday to Tuesday and state whether the balance increased or decreased. b) Identify the lowest balance and write an absolute-difference expression for its distance from Friday's balance. c) Write the four absolute daily changes between consecutive days and identify the greatest one.

Hints

- Keep signed change and magnitude of change distinct. - Use absolute difference when comparing how large two balances or daily changes are. - In part c), calculate all four consecutive-day magnitudes before choosing the greatest.

Solution

1. a) \(-4-12=-16\), so the balance decreased by \(\$16\). 2. b) The lowest balance is \(-\$18\) on Wednesday. Its distance from Friday is \(|22-(-18)|=40\), or \(\$40\). 3. c) The absolute daily changes are \(|-4-12|=16\), \(|-18-(-4)|=14\), \(|-9-(-18)|=9\), and \(|22-(-9)|=31\). The greatest is \(\$31\) from Thursday to Friday.

Answer

a) \(-4-12=-16\); the balance decreased by \(\$16\). b) Wednesday; \(|22-(-18)|=\$40\). c) Daily changes: \(\$16\), \(\$14\), \(\$9\), \(\$31\); greatest from Thursday to Friday.
5185777
Which tasks have equal values? Sort the letters into the groups with values \(-25\) and \(-55\). (A) \(-40+15\) (B) \(-40-(-15)\) (C) Subtract \(15\) from \(-40\). (D) Add \(-15\) to \(-40\). (E) \(-40-15\) (F) Add \(15\) to \(-40\). (G) Subtract \(-15\) from \(-40\).

Hints

- Translate each verbal description into an expression. - Subtracting a negative number is the same as adding a positive number. - Evaluate each task before grouping the letters.

Solution

1. A, B, F, and G each equal \(-25\). 2. C, D, and E each equal \(-55\). 3. “Subtract \(x\) from \(y\)” means \(y-x\), while “add \(x\) to \(y\)” means \(y+x\).

Answer

Value \(-25\): A, B, F, G Value \(-55\): C, D, E
5186327
Find the sum \(1 + 2 + 3 + \dots + 79 + 80\) by pairing terms. 1) Name the two properties that let you reorder and regroup the addends. 2) How many pairs are formed when you pair the first term with the last term, the second term with the next-to-last term, and so on? 3) What is the sum of each pair? 4) Find the total sum.

Hints

- Compare the sums of the first and last terms and of the second and next-to-last terms. - Divide the number of terms by \(2\) to find the number of pairs. - Multiply the number of pairs by the sum of each pair.

Solution

1. The commutative property allows the terms to be reordered, and the associative property allows them to be regrouped. 2. There are \(80 \div 2 = 40\) pairs. 3. Each pair has a sum of \(81\), since \(1 + 80 = 81\), \(2 + 79 = 81\), and so on. 4. Therefore, the total is \(40 \cdot 81 = 3240\).

Answer

1) Commutative property and associative property 2) \(40\) pairs 3) \(81\) 4) \(3240\)
5186337
An apple seller stacks apples in a triangular display. The top row has \(1\) apple, and each row below it has one more apple than the row above. The bottom row has \(40\) apples. 1) Write an expression for the total number of apples. 2) Find the total efficiently by pairing terms. 3) Explain why \(20 \cdot 41\) gives the total.

Hints

- Write the first few and last few terms of the sum. - Pair terms from opposite ends of the list. - Determine how many pairs are formed and the sum of each pair.

Solution

1. The total is represented by \(1 + 2 + 3 + \dots + 40\). 2. Pair the first and last terms: \(1 + 40 = 41\), \(2 + 39 = 41\), and so on. Since \(40 \div 2 = 20\), there are \(20\) pairs. 3. Each pair has a sum of \(41\), so the total is \(20 \cdot 41 = 820\).

Answer

1) \(1 + 2 + 3 + \dots + 40\) 2) \(820\) apples 3) There are \(20\) pairs, and each pair has a sum of \(41\).
5186347
Find the sum of all even numbers from \(2\) through \(60\): \(2 + 4 + 6 + \dots + 58 + 60\) 1) How many terms are in the sum? 2) Pair terms that have the same sum. How many pairs are there, and what is the sum of each pair? 3) Find the total.

Hints

- Divide \(60\) by \(2\) to count the even numbers. - Pair the smallest term with the largest term. - Multiply the number of pairs by the sum of each pair.

Solution

1. There are \(60 \div 2 = 30\) even numbers from \(2\) through \(60\). 2. Pair terms from opposite ends: \(2 + 60 = 62\), \(4 + 58 = 62\), and so on. The \(30\) terms form \(30 \div 2 = 15\) pairs. 3. The total is \(15 \cdot 62 = 930\).

Answer

1) \(30\) terms 2) \(15\) pairs; each pair has a sum of \(62\) 3) \(930\)
5186597
Evaluate each expression. a) \(-12{,}450+5670\) b) \(34{,}200-(-8950)\) c) \(-7300-12{,}800\) d) \(10^4-100{,}000\)

Hints

- Distinguish signs attached to numbers from operation signs. - Subtracting a negative number becomes addition. - Evaluate the power of ten before subtracting.

Solution

1. In a), \(-12{,}450+5670=-6780\). 2. In b), \(34{,}200-(-8950)=34{,}200+8950=43{,}150\). 3. In c), \(-7300-12{,}800=-20{,}100\). 4. In d), \(10^4=10{,}000\), so \(10{,}000-100{,}000=-90{,}000\).

Answer

a) \(-6780\) b) \(43{,}150\) c) \(-20{,}100\) d) \(-90{,}000\)
5186607
Insert \(<\), \(>\), or \(=\) to make each statement true. a) \(-3500+1200\mathbin{\square}-3500-1200\) b) \(-800-450\mathbin{\square}-800-(-450)\) c) \(15{,}000-25{,}000\mathbin{\square}-5000-6000\) d) \(-10^3+1\mathbin{\square}0\)

Hints

- Evaluate both sides separately. - For negative numbers, the value with the greater absolute value is smaller. - Compare the results on a number line.

Solution

1. In a), the values are \(-2300\) and \(-4700\), so use \(>\). 2. In b), the values are \(-1250\) and \(-350\), so use \(<\). 3. In c), the values are \(-10{,}000\) and \(-11{,}000\), so use \(>\). 4. In d), \(-10^3+1=-999\), so use \(<\).

Answer

a) \(>\) b) \(<\) c) \(>\) d) \(<\)
5186817
Write an expression and evaluate it. From the sum of \(-412\) and \(187\), subtract the difference of \(-256\) and \(-58\).

Hints

- Translate “sum” and “difference” into operations. - Use parentheses to keep the two intermediate expressions separate. - The wording “from the sum, subtract the difference” determines the order.

Solution

1. The sum is \(-412+187=-225\). 2. The difference is \(-256-(-58)=-198\). 3. Subtract the results: \(-225-(-198)=-27\).

Answer

\((-412+187)-(-256-(-58))=-27\)
5186887
Leon claims, “When I subtract a negative number from another negative number, the result is always negative.” Disprove his claim with two examples: 1. One example with a positive result. 2. One example with a result of \(0\). Evaluate both differences.

Hints

- Rewrite subtraction of a negative number as addition. - Try negative numbers with different absolute values. - A number minus itself equals zero.

Solution

1. For a positive result, the number being subtracted can have a greater absolute value than the starting number. For example, \((-2)-(-5)=3\). 2. For a result of zero, subtract a negative number from itself. For example, \((-8)-(-8)=0\). 3. Since neither result is negative, the examples disprove the claim.

Answer

Possible examples are: 1. \((-2)-(-5)=3\) 2. \((-8)-(-8)=0\)
5186897
Consider \((-14)-(-20)\). a) Evaluate the difference. b) Explain why the result is positive even though both numbers in the expression are negative.

Hints

- Rewrite subtraction of a negative as addition. - Picture the calculation as movement on a number line. - Compare the absolute values of \(-14\) and \(20\).

Solution

1. Rewrite the subtraction: \((-14)-(-20)=-14+20\). 2. Evaluate: \(-14+20=6\). 3. The result is positive because adding \(20\) moves \(20\) units right from \(-14\), crossing zero and ending at \(6\).

Answer

a) \(6\) b) Subtracting \(-20\) is the same as adding \(20\). Since \(20>14\), the result is positive.
5186907
Use the numbers \(-6\) and \(-10\). 1. Evaluate \((-6)-(-10)\). 2. Evaluate \((-10)-(-6)\). Compare the results. What do you notice about their signs?

Hints

- Rewrite each subtraction of a negative number as addition. - Compare both the signs and absolute values of the results. - Think about what happens when the order in a subtraction is reversed.

Solution

1. \((-6)-(-10)=-6+10=4\). 2. \((-10)-(-6)=-10+6=-4\). 3. The results have the same absolute value but opposite signs. Reversing the order of a subtraction negates the difference.

Answer

1. \(4\) 2. \(-4\) The results have equal absolute values and opposite signs.
5187027
Write an expression for the description and evaluate it. Subtract \(-150\) from the sum of \(-65\) and \(38\).

Hints

- Translate “sum” into addition. - The wording “subtract \(-150\) from” determines the order. - Subtracting a negative number is the same as adding a positive number.

Solution

1. The expression is \((-65+38)-(-150)\). 2. The sum is \(-65+38=-27\). 3. Then \(-27-(-150)=-27+150=123\).

Answer

\((-65+38)-(-150)=123\)
5187037
Write an expression for the description and evaluate it. Add the difference of \(16\) and \(-34\) to the difference of \(210\) and \(450\).

Hints

- Separate the description into two differences. - “The difference of \(a\) and \(b\)” means \(a-b\). - Evaluate each grouped difference before adding them.

Solution

1. The expression is \((210-450)+[16-(-34)]\). 2. The first difference is \(210-450=-240\). 3. The second difference is \(16-(-34)=50\). 4. Add the results: \(-240+50=-190\).

Answer

\((210-450)+[16-(-34)]=-190\)
5187147
Rewrite the expression to make it easy to evaluate. Keep each operation sign with its number when you reorder the terms. \(864 + 257 - 364 + 43\)

Hints

- Look for two numbers with matching final digits that make an easy difference. - Look for two addends that make a multiple of \(100\). - When reordering, move each operation sign with the term that follows it.

Solution

1. Reorder the terms: \(864 - 364 + 257 + 43\). 2. Group convenient pairs: \((864 - 364) + (257 + 43)\). 3. Evaluate: \(500 + 300 = 800\).

Answer

\(800\)
5187517
Evaluate \(30-[(-12+25)-(8-15)]\).

Hints

- Work from the innermost grouping symbols outward. - Write each intermediate value to keep track of the signs. - Subtracting a negative number is the same as adding a positive number.

Solution

1. Evaluate the inner parentheses: \(-12+25=13\) and \(8-15=-7\). 2. Evaluate the brackets: \(13-(-7)=20\). 3. Complete the calculation: \(30-20=10\).

Answer

\(10\)
5188967
Evaluate each expression using the order of operations. a) \(84-(112-250)\) b) \(-37+(-44+19)+60\) c) \(-(55-92)-(18+24)\)

Hints

- Evaluate parentheses first. - A negative sign outside parentheses takes the opposite of the parenthetical value. - Write each intermediate value before continuing.

Solution

1. In a), \(112-250=-138\), so \(84-(-138)=222\). 2. In b), \(-44+19=-25\), so \(-37-25+60=-2\). 3. In c), \(55-92=-37\) and \(18+24=42\), so \(-(-37)-42=-5\).

Answer

a) \(222\) b) \(-2\) c) \(-5\)
5188977
Evaluate each expression with nested grouping symbols. a) \(15-[32-(14-50)]\) b) \(-21+[(-45+18)-(-12)]\)

Hints

- Work from the innermost grouping symbols outward. - Record each intermediate result. - Pay close attention when subtracting a negative value.

Solution

1. In a), \(14-50=-36\). Then \(32-(-36)=68\), and \(15-68=-53\). 2. In b), \(-45+18=-27\). Then \(-27-(-12)=-15\), and \(-21+(-15)=-36\).

Answer

a) \(-53\) b) \(-36\)
5189537
Evaluate \(-35-(15-60)+(-20)\).

Hints

- Evaluate the parentheses first. - Subtracting a negative number becomes addition. - Work one operation at a time.

Solution

1. Evaluate inside the parentheses: \(15-60=-45\). 2. Substitute the result: \(-35-(-45)+(-20)\). 3. Evaluate: \(-35+45-20=-10\).

Answer

\(-10\)
5190117
Evaluate \((120-250)+(85-(-15))\) step by step.

Hints

- Evaluate the two parenthetical expressions separately. - Subtracting a negative number becomes addition. - Add the two intermediate results with their signs.

Solution

1. Evaluate the first parentheses: \(120-250=-130\). 2. Evaluate the second parentheses: \(85-(-15)=100\). 3. Add the results: \(-130+100=-30\).

Answer

\(-30\)
5195987
Rewrite the expression to make it easy to evaluate. Keep each operation sign with its number when reordering the terms. \(45 - 17 + 55 - 83 + 12\)

Hints

- Group terms to create multiples of \(100\). - Keep each subtraction sign with the number that follows it. - Find the total of the positive terms and the total being subtracted.

Solution

1. Group the positive terms and the amounts being subtracted: \((45 + 55 + 12) - (17 + 83)\). 2. Evaluate the groups: \(112 - 100 = 12\).

Answer

\(12\)
5199567
A checking account begins with a balance of \(\$120\). During the morning, \(\$150\) is withdrawn for a purchase, and then another \(\$40\) is withdrawn at an ATM. Use subtraction expressions to find the balance after each withdrawal.

Hints

- A withdrawal decreases the balance. - Find the balance after the first transaction before applying the second. - Spending more money when the balance is already negative makes the balance more negative.

Solution

1. After the purchase, \(120-150=-30\), so the balance is \(-\$30\). 2. After the ATM withdrawal, \(-30-40=-70\), so the balance is \(-\$70\).

Answer

After the purchase: \(\$120-\$150=-\$30\) After the ATM withdrawal: \(-\$30-\$40=-\$70\)
5204397
Use the numbers \(-12,7,-3,0,-8,2\). a) Order the numbers from greatest to least using \(>\). b) Find the pair of numbers with the least sum. c) Find the two numbers that are farthest apart on a number line.

Hints

- Picture the numbers on a number line before ordering them. - To make the least sum, consider the two least values. - The greatest distance occurs between the least and greatest values.

Solution

1. Ordering the numbers from greatest to least gives \(7>2>0>-3>-8>-12\). 2. The least sum comes from the two least numbers: \(-12+(-8)=-20\). The pair is \(-12\) and \(-8\). 3. The greatest distance is between the least and greatest numbers: \(7-(-12)=19\). The pair is \(-12\) and \(7\).

Answer

a) \(7>2>0>-3>-8>-12\) b) \(-12\) and \(-8\) c) \(-12\) and \(7\)
5204407
Use the numbers \(-10,4,-1,6,-5,3\). a) Which numbers lie strictly between \(-6\) and \(4\) on a number line? b) Order all six numbers from least to greatest using \(<\). c) Find a pair of numbers whose sum is \(5\).

Hints

- “Strictly between” means that the endpoints are not included. - Place the values mentally on a number line to order them. - For the sum, test pairs containing one positive and one negative number.

Solution

1. The numbers greater than \(-6\) and less than \(4\) are \(-5,-1,3\). 2. From least to greatest, the numbers are \(-10<-5<-1<3<4<6\). 3. Since \(6+(-1)=5\), one qualifying pair is \(6\) and \(-1\).

Answer

a) \(-5,-1,3\) b) \(-10<-5<-1<3<4<6\) c) \(6\) and \(-1\)
5204697
A diving robot begins at \(-45\,\text{m}\) relative to sea level and descends \(15\,\text{m}\) each minute. After how many whole minutes will it first be below \(-110\,\text{m}\)?

Hints

- Record the robot's position after each minute. - “Below \(-110\,\text{m}\)” means the position must be less than \(-110\). - How does each descent change the signed position?

Solution

1. Subtract \(15\) for each minute: after \(1\) minute the position is \(-60\,\text{m}\), after \(2\) minutes it is \(-75\,\text{m}\), after \(3\) minutes it is \(-90\,\text{m}\), and after \(4\) minutes it is \(-105\,\text{m}\). 2. After \(5\) minutes, the position is \(-105-15=-120\,\text{m}\). 3. Since \(-120<-110\) and the position after \(4\) minutes is not below \(-110\), the first such time is \(5\) minutes.

Answer

The robot will first be below \(-110\,\text{m}\) after \(5\) minutes.
5215337
Estimate by rounding to the nearest ten. Then evaluate \(128-57-43+72-110\) exactly using efficient regrouping.

Hints

- Round each term to the nearest ten. - Look for pairs that make \(200\) and \(-100\). - Compare the exact value with the estimate.

Solution

1. The estimate is \(130-60-40+70-110=-10\). 2. Regroup the exact expression as \((128+72)+(-57-43)-110\). 3. This gives \(200-100-110=-10\).

Answer

Estimate: approximately \(-10\) Exact value: \(-10\)
5216447
Evaluate \(-42+(18-50)+60\).

Hints

- Evaluate the parentheses first. - A positive sign before the parentheses leaves the result unchanged. - Then add from left to right.

Solution

1. Evaluate the parentheses: \(18-50=-32\). 2. Then \(-42+(-32)+60=-74+60=-14\).

Answer

\(-14\)
5216457
Evaluate \(80-[-30-(25-45)]\).

Hints

- Work from the innermost grouping symbols outward. - Pay attention to every subtraction sign before a grouped value. - Subtracting a negative number becomes addition.

Solution

1. Evaluate the inner parentheses: \(25-45=-20\). 2. Evaluate the brackets: \(-30-(-20)=-10\). 3. Complete the calculation: \(80-(-10)=90\).

Answer

\(90\)
5217597
Evaluate the four expressions, and then order their values from least to greatest. a) \(-110+(-40)\) b) \(-110-(-40)\) c) \(110+(-110)\) d) \(0-110\)

Hints

- Evaluate all four expressions first. - Among negative numbers, the value with the greater absolute value is smaller. - Use a number line to order the results.

Solution

1. The values are a) \(-150\), b) \(-70\), c) \(0\), and d) \(-110\). 2. From least to greatest, \(-150<-110<-70<0\). 3. The corresponding order is a), d), b), c).

Answer

a) \(-150\) b) \(-70\) c) \(0\) d) \(-110\) Order: \(-150<-110<-70<0\), or a), d), b), c)
5217707
Rewrite the expression to make it easy to evaluate. Keep each subtraction sign with the number that follows it. \(154 + 78 - 54 - 28\)

Hints

- Look for pairs that make easy differences. - Keep each operation sign with its term when regrouping. - Add the two resulting differences.

Solution

1. Regroup as \((154 - 54) + (78 - 28)\). 2. Evaluate: \(100 + 50 = 150\).

Answer

\((154 - 54) + (78 - 28) = 150\)
5217717
Evaluate \(-123+57+265-77+43-65\) efficiently using properties of addition.

Hints

- Look for pairs that make convenient hundreds. - Keep each sign attached to its number when reordering. - Use the commutative and associative properties.

Solution

1. Reorder and regroup as \((-123-77)+(57+43)+(265-65)\). 2. The partial sums are \(-200\), \(100\), and \(200\). 3. Therefore, the value is \(-200+100+200=100\).

Answer

\(100\)
5217907
Add the difference of \(45\) and \(120\) to the sum of \(-33\) and \(-17\).

Hints

- Translate “difference” and “sum” into operations. - Write the full expression with parentheses. - Evaluate each grouped part first.

Solution

1. The difference is \(45-120=-75\). 2. The sum is \(-33+(-17)=-50\). 3. Adding the results gives \(-75+(-50)=-125\).

Answer

\(-125\)
5217917
Subtract the sum of \(-215\) and \(90\) from the opposite of \(45\).

Hints

- First find the opposite of \(45\). - The wording “subtract the sum from” determines the order. - Subtracting a negative number is the same as adding a positive number.

Solution

1. The opposite of \(45\) is \(-45\). 2. The sum is \(-215+90=-125\). 3. Subtract the sum: \(-45-(-125)=80\).

Answer

\(80\)
5217927
Find the sum of \(-15\), \(42\), and \(-10\). Subtract that sum from the difference of \(-100\) and \(50\).

Hints

- Evaluate the sum and the difference separately. - The final subtraction is difference minus sum. - Subtracting a positive number from a negative number makes the result more negative.

Solution

1. The sum is \(-15+42+(-10)=17\). 2. The difference is \(-100-50=-150\). 3. Subtract the sum from the difference: \(-150-17=-167\).

Answer

\(-167\)
5225607
Ms. Miller begins the week with \(\$85\) in her checking account. During the week, these transactions occur: 1. A \(\$120\) grocery purchase is debited. 2. A credit of \(\$50\) is deposited. 3. A \(\$30\) bill is paid. Calculate the balance after each transaction. What does the sign of the final balance mean?

Hints

- Apply the transactions in the order given. - Purchases and bill payments decrease the balance; credits increase it. - A positive balance represents available money, while a negative balance represents an overdraft.

Solution

1. After the grocery purchase, the balance is \(85-120=-35\), or \(-\$35\). 2. After the credit, the balance is \(-35+50=15\), or \(\$15\). 3. After the bill payment, the balance is \(15-30=-15\), or \(-\$15\). 4. The negative sign means the account is overdrawn by \(\$15\).

Answer

After the first transaction, the balance is \(-\$35\); after the second, it is \(\$15\); and after the third, it is \(-\$15\). The negative sign means the account is overdrawn by \(\$15\).
5226177
A weather station records two temperature changes on each of three days. Find the final evening temperature for each day. 1. Monday begins at \(5\,^\circ\text{C}\). The temperature rises \(8\,^\circ\text{C}\), then falls \(10\,^\circ\text{C}\). 2. Tuesday begins at \(-3\,^\circ\text{C}\). The temperature rises \(6\,^\circ\text{C}\), then falls \(7\,^\circ\text{C}\). 3. Wednesday begins at \(-1\,^\circ\text{C}\). The temperature falls \(4\,^\circ\text{C}\), then rises \(5\,^\circ\text{C}\).

Hints

- Picture the temperatures on a vertical number line. - For each change, decide whether to move up or down. - Apply the changes in order from the starting temperature.

Solution

1. Monday: \(5+8-10=3\), so the final temperature is \(3\,^\circ\text{C}\). 2. Tuesday: \(-3+6-7=-4\), so the final temperature is \(-4\,^\circ\text{C}\). 3. Wednesday: \(-1-4+5=0\), so the final temperature is \(0\,^\circ\text{C}\).

Answer

1. \(3\,^\circ\text{C}\) 2. \(-4\,^\circ\text{C}\) 3. \(0\,^\circ\text{C}\)
5226187
An elevator uses Level \(0\) for street level, positive numbers for levels above street level, and negative numbers for levels below street level. a) The elevator starts on Level \(2\), travels down \(5\) levels, and then travels up \(2\) levels. Where does it stop? b) The elevator is on Level \(-1\) and travels down \(3\) more levels. Where does it stop? c) The elevator is on Level \(-2\) and must travel to Level \(3\). How many levels must it travel upward?

Hints

- Picture the levels on a vertical number line. - Moving up increases the level number, and moving down decreases it. - In part c), find the distance from the starting level to the destination.

Solution

1. For part a), \(2-5+2=-1\), so the elevator stops on Level \(-1\). 2. For part b), \(-1-3=-4\), so the elevator stops on Level \(-4\). 3. For part c), the distance from \(-2\) to \(3\) is \(3-(-2)=5\), so the elevator must travel up \(5\) levels.

Answer

a) Level \(-1\) b) Level \(-4\) c) \(5\) levels upward
5226197
For each movement on a number line, find the missing value and write the corresponding addition equation. a) Start at \(-5\) and move \(8\) units right. b) Start at \(3.5\) and move \(6\) units left. c) You end at \(-2\) after moving \(4\) units left from an unknown starting point. Find the starting point.

Hints

- Moving right corresponds to adding a positive number. - Moving left corresponds to adding a negative number. - To find an unknown starting point, undo the movement.

Solution

1. For a), moving right means adding \(8\): \(-5+8=3\). 2. For b), moving left means adding \(-6\): \(3.5+(-6)=-2.5\). 3. For c), let the starting point be \(s\). Then \(s+(-4)=-2\), so \(s=2\). The complete equation is \(2+(-4)=-2\).

Answer

a) \(3\); \(-5+8=3\) b) \(-2.5\); \(3.5+(-6)=-2.5\) c) \(2\); \(2+(-4)=-2\)
5226207
Analyze sums on a number line. a) Which sum lies farther left: \((-12)+7\) or \((-3)+(-3)\)? Calculate and compare. b) A point is at \(-4.8\). What number \(x\) must be added to move it to \(0\)? c) A point at \(-2.4\) is moved to its opposite by adding a number \(y\). Find \(y\) and write the complete equation.

Hints

- The value farther left on a number line is smaller. - A number plus its opposite equals \(0\). - Find the change needed to move from \(-2.4\) to \(2.4\).

Solution

1. For a), \((-12)+7=-5\) and \((-3)+(-3)=-6\). Since \(-6<-5\), \((-3)+(-3)\) lies farther left. 2. For b), add the opposite of \(-4.8\): \(-4.8+4.8=0\). Therefore, \(x=4.8\). 3. For c), the opposite of \(-2.4\) is \(2.4\). The required change is \(2.4-(-2.4)=4.8\), so \(y=4.8\). The equation is \(-2.4+4.8=2.4\).

Answer

a) \((-3)+(-3)\), because \(-6<-5\) b) \(x=4.8\) c) \(y=4.8\); \(-2.4+4.8=2.4\)
5226317
Complete the table by finding each missing value of \(x\), \(y\), or \(x+y\). <table> <thead> <tr> <th>Item</th> <th>1</th> <th>2</th> <th>3</th> <th>4</th> <th>5</th> </tr> </thead> <tbody> <tr> <td>\(x\)</td> <td>\(-7.2\)</td> <td>\(3 \frac{1}{4}\)</td> <td>\(-0.8\)</td> <td>\(-\frac{5}{6}\)</td> <td>\(-4.15\)</td> </tr> <tr> <td>\(y\)</td> <td>\(4.5\)</td> <td>\(-5 \frac{1}{2}\)</td> <td>\(\dots\)</td> <td>\(\frac{1}{3}\)</td> <td>\(4.15\)</td> </tr> <tr> <td>\(x+y\)</td> <td>\(\dots\)</td> <td>\(\dots\)</td> <td>\(-2\)</td> <td>\(\dots\)</td> <td>\(\dots\)</td> </tr> </tbody> </table>

Hints

- For each column, decide whether you need to add two rational numbers or find a missing addend. - Fractions need a common denominator before they can be added. - Converting between fractions and decimals can make some calculations easier. - When a sum and one addend are known, subtract the known addend to find the other one.

Solution

1. For item 1, \(-7.2+4.5=-2.7\). 2. For item 2, \(3 \frac{1}{4}+\left(-5 \frac{1}{2}\right)=3 \frac{1}{4}-5 \frac{2}{4}=-2 \frac{1}{4}\). 3. For item 3, \(-0.8+y=-2\), so \(y=-2-(-0.8)=-1.2\). 4. For item 4, \(-\frac{5}{6}+\frac{1}{3}=-\frac{5}{6}+\frac{2}{6}=-\frac{1}{2}\). 5. For item 5, \(-4.15+4.15=0\).

Answer

1) \(-2.7\) 2) \(-2 \frac{1}{4}\) 3) \(y=-1.2\) 4) \(-\frac{1}{2}\) 5) \(0\)
5226327
Evaluate each expression, then order the results from least to greatest. a) \(-12.5+7.8\) b) \(-\frac{3}{5}-\frac{1}{2}\) c) \(4 \frac{1}{3}+\left(-2 \frac{5}{6}\right)\) d) \(-0.75-(-1.2)\)

Hints

- Evaluate each expression before comparing the results. - Be careful when subtracting a negative number. - Converting all results to the same form can make them easier to compare.

Solution

1. For a), \(-12.5+7.8=-4.7\). 2. For b), \(-\frac{3}{5}-\frac{1}{2}=-\frac{6}{10}-\frac{5}{10}=-\frac{11}{10}=-1.1\). 3. For c), \(4 \frac{1}{3}+\left(-2 \frac{5}{6}\right)=\frac{13}{3}-\frac{17}{6}=\frac{26}{6}-\frac{17}{6}=\frac{9}{6}=1.5\). 4. For d), \(-0.75-(-1.2)=-0.75+1.2=0.45\). 5. Compare the results: \(-4.7<-1.1<0.45<1.5\). Therefore, the order is \(a<b<d<c\).

Answer

a) \(-4.7\) b) \(-1.1\) c) \(1.5\) d) \(0.45\) Order: \(a<b<d<c\)
5226357
Evaluate each sum using the commutative and associative properties. For every part, write the regrouped expression you use before giving the result. 1) \((-24) + 15 + (-16) + 35\) 2) \(6.7 + (-3.2) + 1.3 + (-1.8)\) 3) \(-\frac{3}{4} + 2\frac{1}{5} - 0.25 + 0.8\)

Hints

- Keep each signed term intact while changing order. - Search for pairings that produce whole numbers. - Convert a rational number only when the conversion creates a useful pair.

Solution

1. \((-24-16)+(15+35)=-40+50=10\). 2. \((6.7+1.3)+(-3.2-1.8)=8-5=3\). 3. Convert \(-\frac{3}{4}=-0.75\), then \((-0.75-0.25)+(2.2+0.8)=-1+3=2\).

Answer

1) \((-24-16)+(15+35)=10\) 2) \((6.7+1.3)+(-3.2-1.8)=3\) 3) \((-0.75-0.25)+(2.2+0.8)=2\)
5226367
Evaluate each sum by choosing and showing pairings that avoid converting every term to one representation. 1) \(0.375 - 5\frac{1}{2} + \frac{5}{8} + 4.5\) 2) \(-8.1 + 2\frac{3}{7} + 1.1 + 5\frac{4}{7}\) 3) \(11\frac{1}{9} - 4.9 - 2\frac{1}{9} - 5.1\)

Hints

- Look for compatible values already in the same representation. - Convert only the term needed to form a particularly simple pair. - Show the pairings in your answer.

Solution

1. Use \(0.375=\frac{3}{8}\): \(\left(\frac{3}{8}+\frac{5}{8}\right)+(-5.5+4.5)=1-1=0\). 2. \((-8.1+1.1)+\left(2\frac{3}{7}+5\frac{4}{7}\right)=-7+8=1\). 3. \(\left(11\frac{1}{9}-2\frac{1}{9}\right)+(-4.9-5.1)=9-10=-1\).

Answer

1) \(\left(\frac{3}{8}+\frac{5}{8}\right)+(-5.5+4.5)=0\) 2) \((-8.1+1.1)+\left(2\frac{3}{7}+5\frac{4}{7}\right)=1\) 3) \(\left(11\frac{1}{9}-2\frac{1}{9}\right)+(-4.9-5.1)=-1\)
5226427
A small snack stand records its daily revenue and expenses. The daily net is \(\text{revenue}-\text{expenses}\). A positive result is a profit, and a negative result is a loss. a) Calculate the daily net for each day. b) Find the total net for all four days. <table> <thead> <tr> <th>Day</th> <th>Revenue</th> <th>Expenses</th> <th>Daily net</th> </tr> </thead> <tbody> <tr> <td>Monday</td> <td>\(\$145\)</td> <td>\(\$162\)</td> <td></td> </tr> <tr> <td>Tuesday</td> <td>\(\$210\)</td> <td>\(\$185\)</td> <td></td> </tr> <tr> <td>Wednesday</td> <td>\(\$95\)</td> <td>\(\$130\)</td> <td></td> </tr> <tr> <td>Thursday</td> <td>\(\$178\)</td> <td>\(\$178\)</td> <td></td> </tr> </tbody> </table>

Hints

- What sign should the net have when expenses are greater than revenue? - Calculate the difference for each day first. - Then add all four signed daily results.

Solution

1. Monday: \(145-162=-17\), so the daily net is \(-\$17\). 2. Tuesday: \(210-185=25\), so the daily net is \(\$25\). 3. Wednesday: \(95-130=-35\), so the daily net is \(-\$35\). 4. Thursday: \(178-178=0\), so the daily net is \(\$0\). 5. The total net is \(-17+25-35+0=-27\), or \(-\$27\).

Answer

a) Monday: \(-\$17\); Tuesday: \(\$25\); Wednesday: \(-\$35\); Thursday: \(\$0\) b) The total net is \(-\$27\).
5226477
At \(6{:}00\) a.m. on a winter day, the temperature is \(-6\,^\circ\text{F}\). a) By noon, the temperature rises \(11\,^\circ\text{F}\). What is the noon temperature? b) By late evening, the temperature drops \(8\,^\circ\text{F}\) from the noon value. What is the evening temperature? c) Find the difference between the highest and lowest of the morning, noon, and evening temperatures.

Hints

- Picture the temperatures on a vertical number line. - Translate “rises” and “drops” into operations. - To find the difference between the highest and lowest values, find their distance on the number line. - Compare all three temperatures before answering part c).

Solution

1. The noon temperature is \(-6+11=5\,^\circ\text{F}\). 2. The evening temperature is \(5-8=-3\,^\circ\text{F}\). 3. The highest temperature is \(5\,^\circ\text{F}\), and the lowest is \(-6\,^\circ\text{F}\). Their difference is \(5-(-6)=11\,^\circ\text{F}\).

Answer

a) \(5\,^\circ\text{F}\) b) \(-3\,^\circ\text{F}\) c) \(11\,^\circ\text{F}\)
5226487
Julia and Tom compare their checking account balances. - Julia has \(\$12\), so her balance is \(+\$12\). - Tom's account is overdrawn by \(\$8\), so his balance is \(-\$8\). a) What is the difference between their current balances? b) Julia buys a book for \(\$15\). What is her new balance? c) Tom deposits \(\$10\). What is his new balance? d) After these changes, who has the greater balance, and what is the difference between the balances?

Hints

- A negative balance is less than a positive balance. - A purchase decreases a balance, while a deposit increases it. - A difference between balances is their distance on a number line. - Subtracting a negative number is equivalent to adding its opposite.

Solution

1. The initial difference is \(12-(-8)=20\), or \(\$20\). 2. Julia's new balance is \(12-15=-3\), or \(-\$3\). 3. Tom's new balance is \(-8+10=2\), or \(\$2\). 4. Since \(2>-3\), Tom has the greater balance. The new difference is \(2-(-3)=5\), or \(\$5\).

Answer

a) \(\$20\) b) \(-\$3\) c) \(\$2\) d) Tom has the greater balance, and the difference is \(\$5\).
5226597
Find each requested value. a) Evaluate \((+7)-(-15)\). b) What number must be subtracted from \(-3.8\) to get \(-10\)? c) Subtract \(+1 \frac{3}{4}\) from \(-2 \frac{1}{2}\).

Hints

- Subtracting a negative number is the same as adding its opposite. - For b), represent the unknown number with a variable or a blank. - For mixed numbers, a common denominator can make the subtraction easier.

Solution

1. For a), subtracting a negative is addition: \(7-(-15)=7+15=22\). 2. For b), let the missing number be \(x\). Then \(-3.8-x=-10\), so \(x=-3.8-(-10)=6.2\). 3. For c), use a common denominator: \(-2 \frac{1}{2}-1 \frac{3}{4}=-\frac{10}{4}-\frac{7}{4}=-\frac{17}{4}=-4 \frac{1}{4}\).

Answer

a) \(22\) b) \(6.2\) c) \(-4 \frac{1}{4}\), or \(-4.25\)
5226607
Explore how changing the order affects subtraction. Let \(x=-6.2\) and \(y=-3.4\). a) Find \(x-y\). b) Find \(y-x\). c) Compare your answers from a) and b). What do you notice? d) A student claims, “When you subtract one negative number from another negative number, the result is always positive.” Give a counterexample to show that the claim is false.

Hints

- Substitute the negative values carefully and pay attention to both operation signs and number signs. - Compare the absolute values and signs of the results in a) and b). - One counterexample is enough to disprove a statement that claims something is always true.

Solution

1. For a), \(x-y=-6.2-(-3.4)=-6.2+3.4=-2.8\). 2. For b), \(y-x=-3.4-(-6.2)=-3.4+6.2=2.8\). 3. The two results are opposites: they have the same absolute value and opposite signs. 4. One counterexample is \((-10)-(-2)=-10+2=-8\). Since the result is negative, the claim is false.

Answer

a) \(-2.8\) b) \(2.8\) c) The results are opposites. d) Example: \((-10)-(-2)=-8\). The result is negative, so the claim is false.
5226627
Evaluate each expression. Write each result as a simplified fraction or mixed number. 1) \(\left(-\frac{7}{8}\right)-\left(+\frac{1}{8}\right)-(-3)\) 2) \(\left(+\frac{2}{5}\right)-\left(-\frac{1}{2}\right)-\left(+\frac{9}{10}\right)\) 3) \(\left(-2 \frac{1}{4}\right)-\left(-\frac{3}{4}\right)-(+1)\)

Hints

- Subtracting a number is the same as adding its opposite. - Check whether the fractions already have a common denominator before finding a new one. - You can write whole numbers as fractions when that helps combine terms. - Converting mixed numbers to improper fractions can make some calculations easier.

Solution

1. For 1), rewrite subtraction of a negative as addition: \(-\frac{7}{8}-\frac{1}{8}+3=-1+3=2\). 2. For 2), use denominator \(10\): \(\frac{4}{10}+\frac{5}{10}-\frac{9}{10}=0\). 3. For 3), \(-2 \frac{1}{4}+\frac{3}{4}-1=-1 \frac{1}{2}-1=-2 \frac{1}{2}\).

Answer

1) \(2\) 2) \(0\) 3) \(-2 \frac{1}{2}\)
5226707
Let \(a\) and \(b\) be positive rational numbers. a) What numbers must be included with the set of positive rational numbers so that \(a-b\) is always in the resulting set, no matter which positive rational numbers are chosen? b) State the condition on \(a\) and \(b\) that makes \(a-b\): 1. equal to \(0\). 2. negative. 3. positive. c) Give one subtraction example with fractions or decimals for each case in part b).

Hints

- Compare the minuend \(a\) with the subtrahend \(b\). - Think about the number that separates positive and negative numbers on a number line. - Choose simple fractions or decimals whose differences you can check exactly.

Solution

1. Include \(0\) and all negative rational numbers. The resulting set is the set of all rational numbers. 2. The difference is \(0\) when \(a=b\), negative when \(a<b\), and positive when \(a>b\). 3. Possible examples are \(1.5-1.5=0\), \(0.2-0.7=-0.5\), and \(5.8-2.3=3.5\).

Answer

a) Include \(0\) and all negative rational numbers. b) 1. \(a=b\) 2. \(a<b\) 3. \(a>b\) c) Answers will vary. One possible set is \(1.5-1.5=0\), \(0.2-0.7=-0.5\), and \(5.8-2.3=3.5\).
5226857
The expressions are \(A=a-b+c\) and \(B=a-(b+c)\). Evaluate and compare them for each assignment. a) \(a=10\), \(b=-4\), \(c=2\) b) \(a=-3.5\), \(b=1.5\), \(c=-5\)

Hints

- Use parentheses when substituting negative values. - Evaluate the parentheses in \(B\) before subtracting. - Compare the final values on a number line.

Solution

1. For a), \(A=10-(-4)+2=16\), while \(B=10-((-4)+2)=12\). Thus, \(A>B\). 2. For b), \(A=-3.5-1.5+(-5)=-10\), while \(B=-3.5-(1.5+(-5))=0\). Thus, \(A<B\).

Answer

a) \(A=16\), \(B=12\), so \(A>B\). b) \(A=-10\), \(B=0\), so \(A<B\).
5226867
Evaluate \(T=-x-(y-z)\) for each set of rational numbers. a) \(x=\frac{2}{3}\), \(y=-\frac{1}{6}\), \(z=\frac{1}{2}\) b) \(x=-2\frac{1}{4}\), \(y=1\frac{3}{8}\), \(z=-0.5\)

Hints

- Convert mixed numbers and decimals to fractions. - Use common denominators for addition and subtraction. - Pay close attention to the negative sign in front of \(x\). - Work inside the parentheses first.

Solution

1. For part a), \(T=-\frac{2}{3}-\left(-\frac{1}{6}-\frac{1}{2}\right)=-\frac{2}{3}-\left(-\frac{2}{3}\right)=0\). 2. For part b), convert the values to fractions: \(x=-\frac{9}{4}\), \(y=\frac{11}{8}\), and \(z=-\frac{1}{2}\). 3. Then \(T=-\left(-\frac{9}{4}\right)-\left(\frac{11}{8}-\left(-\frac{1}{2}\right)\right)=\frac{18}{8}-\frac{15}{8}=\frac{3}{8}\).

Answer

a) \(0\) b) \(\frac{3}{8}\)
5227017
A weather station recorded these temperatures at midnight during one week: Monday: \(-4\,^\circ\text{C}\), Tuesday: \(-1\,^\circ\text{C}\), Wednesday: \(3\,^\circ\text{C}\), Thursday: \(5\,^\circ\text{C}\), Friday: \(-2\,^\circ\text{C}\), Saturday: \(-6\,^\circ\text{C}\), Sunday: \(0\,^\circ\text{C}\). a) Find the highest and lowest temperatures. b) Find the difference between the warmest and coldest temperatures. c) List the days from coldest to warmest.

Hints

- Compare the temperatures by placing them on a number line. - To find the difference, subtract the lower temperature from the higher temperature. - Order negative temperatures carefully: a value farther below zero is colder.

Solution

1. The lowest temperature is \(-6\,^\circ\text{C}\) on Saturday, and the highest is \(5\,^\circ\text{C}\) on Thursday. 2. The temperature difference is \(5-(-6)=11\), so it is \(11\,^\circ\text{C}\). 3. Ordering the temperatures gives Saturday, Monday, Friday, Tuesday, Sunday, Wednesday, Thursday.

Answer

a) Highest: \(5\,^\circ\text{C}\); lowest: \(-6\,^\circ\text{C}\) b) \(11\,^\circ\text{C}\) c) Saturday, Monday, Friday, Tuesday, Sunday, Wednesday, Thursday
5227327
Let \(a\) and \(b\) be rational numbers. a) Under what condition is \(a - b > a\)? Explain why and give a numerical example. b) Under what condition does \(a - b = a + b\)? Justify your answer algebraically.

Hints

- Subtract the common term \(a\) from both sides. - Subtracting a negative number is equivalent to adding a positive number. - In part b), collect all terms containing \(b\) on one side.

Solution

1. For a), subtract \(a\) from both sides of \(a - b > a\): \(-b > 0\). This is equivalent to \(b < 0\). 2. Subtracting a negative number increases the original value. For example, \(7 - (-3) = 10\), and \(10 > 7\). 3. For b), subtract \(a\) from both sides: \(-b = b\). 4. Add \(b\): \(0 = 2b\). Divide by \(2\): \(b = 0\). The value of \(a\) can be any rational number.

Answer

a) \(b < 0\). For example, \(7 - (-3) = 10 > 7\). b) \(b = 0\)
5227427
An elevator labels street level as \(0\), levels above it with positive integers, and parking levels below it with negative integers. A passenger enters on Level \(5\) and makes these trips in order: 1. Down \(7\) levels 2. Up \(3\) levels 3. Down \(4\) levels 4. Up \(2\) levels a) On which level does the passenger end? b) What is the lowest level reached during the entire trip? c) How many levels apart are the highest and lowest points of the trip?

Hints

- Record the current level after each move. - Moving down is a negative change, and moving up is a positive change. - Include the starting level when identifying the highest and lowest points. - Find the distance between the highest and lowest signed values.

Solution

1. Track each position: \(5-7=-2\), \(-2+3=1\), \(1-4=-3\), and \(-3+2=-1\). The final position is Level \(-1\). 2. The visited levels are \(5,-2,1,-3,-1\). The least value is \(-3\), so the lowest level reached is Level \(-3\). 3. The highest level is \(5\), and the lowest is \(-3\). Their distance is \(5-(-3)=8\) levels.

Answer

a) Level \(-1\) b) Level \(-3\) c) \(8\) levels
5244887
Analyze each statement about rational-number operations. a) Can \(x-y\) be greater than \(x\)? Explain or give an example. b) If \(a\ne0\) and \(b\ne0\), can \(a\div b\) equal \(0\)? Explain. c) Is the opposite \(-x\) always less than \(x\)? Consider positive values, negative values, and zero.

Hints

- Think about what happens when you subtract a negative number. - Recall when a fraction or quotient equals zero. - Test the statement using a positive number, a negative number, and zero.

Solution

1. The inequality \(x-y>x\) simplifies to \(-y>0\), so \(y<0\). Thus the difference is greater than \(x\) when the subtrahend is negative. For example, \(5-(-2)=7>5\). 2. A quotient with a nonzero divisor equals \(0\) only when its dividend is \(0\). Since \(a\ne0\), \(a\div b\) cannot equal \(0\). 3. If \(x>0\), then \(-x<x\). If \(x<0\), then \(-x>x\). If \(x=0\), then \(-x=x\). Therefore, \(-x<x\) only when \(x\) is positive.

Answer

a) Yes; this happens when \(y<0\). For example, \(5-(-2)=7\). b) No; a quotient equals \(0\) only when the dividend is \(0\). c) No. The statement is true only when \(x>0\).
5318087
Use the number line. a) Write the value of each point \(A\), \(B\), and \(C\) as both a decimal and a fraction in simplest form or a mixed number. b) Find the distance from \(A\) to \(B\) and from \(B\) to \(C\).
Figure for problem 531808

Hints

- Determine the value of one small interval. - Use the point’s position relative to \(0\) to determine its sign. - Distance is the absolute value of the difference between two coordinates. - Convert quarters to decimals or decimals to fractions as needed.

Solution

1. There are \(4\) equal intervals between consecutive integers, so each interval represents \(0.25=\frac{1}{4}\). 2. Point \(A\) is at \(-1.75=-\frac{7}{4}=-1\frac{3}{4}\). 3. Point \(B\) is at \(-0.5=-\frac{1}{2}\). 4. Point \(C\) is at \(0.25=\frac{1}{4}\). 5. The distance from \(A\) to \(B\) is \(\left|-0.5-(-1.75)\right|=1.25=\frac{5}{4}\). 6. The distance from \(B\) to \(C\) is \(|0.25-(-0.5)|=0.75=\frac{3}{4}\).

Answer

a) \(A=-1.75=-\frac{7}{4}\); \(B=-0.5=-\frac{1}{2}\); \(C=0.25=\frac{1}{4}\) b) \(AB=1.25=\frac{5}{4}\); \(BC=0.75=\frac{3}{4}\)
5319147
Operations with decimals can be shown clearly on a number line. Each arrow shows the direction and size of a jump. Find the numbers represented by the red question marks on the number lines.
Figure for problem 531914

Hints

- Pay close attention to the signs of both the jumps and the numbers on the number line. - When you need to work backward to find a starting value, use the inverse operation. - For several arrows, follow the jumps in the order shown.

Solution

1. For a), start at \(3.5\) and make a jump of \(-4.8\): \(3.5-4.8=-1.3\). 2. For b), the starting value is unknown. Since a jump of \(+2.6\) lands at \(-0.8\), work backward: \(-0.8-2.6=-3.4\). 3. For c), start at \(-1.2\). The first jump gives \(-1.2-1.5=-2.7\). Then \(-2.7+3.1=0.4\).

Answer

a) \(-1.3\) b) \(-3.4\) c) First number: \(-2.7\); second number: \(0.4\)
5319167
These number lines show subtracting positive numbers as jumps to the left. Find the value represented by the question mark in each diagram. a) Where does the jump end? b) Where did the jump begin? c) What is the missing middle value?
Figure for problem 531916

Hints

- In these diagrams only, each leftward jump represents subtracting the positive size named on the arrow. - For a missing start, undo the displayed jump from the known endpoint. - In a chain, use the known start and first jump to locate the missing middle value.

Solution

1. a) \(8-23=-15\). 2. b) If \(x-34=-12\), then \(x=22\). 3. c) \(5-18=-13\); the next jump gives \(-13-27=-40\), matching the displayed endpoint.

Answer

a) \(-15\) b) \(22\) c) \(-13\)
5321867
The plot shows the temperature at a mountain weather station from \(6{:}00\) a.m. to \(4{:}00\) p.m. on a winter day. Three students wrote different expressions to determine the temperature at \(4{:}00\) p.m. Lucas: \(-3-1+2+3+4+2+0-1-3-4-2\) Clara: \(-3+(2+3+4+2)-(1+1+3+4+2)\) Jonas: \(-3-1+11-10\) a) Evaluate each expression. What temperature does each expression give for \(4{:}00\) p.m.? b) Explain how each student grouped or represented the hourly temperature changes shown in the plot.
Figure for problem 532186

Hints

- What does the initial \(-3\) in each expression represent on the plot? - Find the temperature change from one marked hour to the next. - Compare how the three expressions group the same signed changes.

Solution

1. Lucas's expression gives \(-3-1+2+3+4+2+0-1-3-4-2=-3\). 2. Clara's expression gives \(-3+(2+3+4+2)-(1+1+3+4+2)=-3+11-11=-3\). 3. Jonas's expression gives \(-3-1+11-10=-3\). Therefore, all three expressions give a final temperature of \(-3\,^{\circ}\text{C}\). 4. Lucas lists every hourly change in order. Clara groups all increases together and all decreases together. Jonas groups consecutive phases: an initial \(1\)-degree decrease, an \(11\)-degree increase, no change, and a final \(10\)-degree decrease.

Answer

a) Each expression equals \(-3\), so the temperature at \(4{:}00\) p.m. is \(-3\,^{\circ}\text{C}\). b) Lucas uses each hourly change. Clara groups positive changes and negative changes. Jonas combines consecutive changes into longer warming and cooling phases.
5331927
The plot shows the air temperature measured at the same time on seven consecutive days at a mountain weather station. Two meteorologists summarize the week: Mr. Frost writes \(T_1=-4+2+3-1-3+2+3\). Ms. Degree writes \(T_2=(2+3+2+3)-(1+3)\). Evaluate both expressions and explain what information each result gives about the weather station.
Figure for problem 533192

Hints

- Start with the point for Day 1. - Relate positive and negative numbers to increases and decreases between marked days. - Compare \(T_1\) with the final plotted temperature and \(T_2\) with the overall change.

Solution

1. For \(T_1\), \(-4+2+3-1-3+2+3=2\). The expression starts with the Day 1 temperature and applies each daily change, so \(2\,^{\circ}\text{C}\) is the temperature on Day 7. 2. For \(T_2\), \((2+3+2+3)-(1+3)=10-4=6\). The first group is the total of all increases, and the second group is the total of all decreases. The result \(6\,^{\circ}\text{C}\) is the net temperature change from Day 1 to Day 7.

Answer

\(T_1=2\,^{\circ}\text{C}\); this is the temperature on Day 7. \(T_2=6\,^{\circ}\text{C}\); this is the net temperature change from Day 1 to Day 7.
5350927
A hiking group is planning a route through the mountains. The bar graph shows the elevation of each stop in feet. a) Read the elevations of the six stops from the graph. b) Find the elevation change between each pair of consecutive stops. Use a negative sign for a descent. c) Find the sum of all the elevation changes from part b). d) A hiker says, “The finish is only \(300\,\text{ft}\) above the start, so we climb only \(300\,\text{ft}\) in all.” Explain why this statement is incorrect.
Figure for problem 535092

Hints

- Read each bar using the y-axis scale. - Subtract each elevation from the next elevation in route order. - Add the signed changes for the net change. - Add only positive changes to find total ascent. - Descents affect net change but not the amount already climbed.

Solution

1. The elevations are Start, \(600\,\text{ft}\); A, \(1500\,\text{ft}\); B, \(2600\,\text{ft}\); C, \(1400\,\text{ft}\); D, \(2300\,\text{ft}\); Finish, \(900\,\text{ft}\). 2. The consecutive changes are \(1500-600=+900\,\text{ft}\), \(2600-1500=+1100\,\text{ft}\), \(1400-2600=-1200\,\text{ft}\), \(2300-1400=+900\,\text{ft}\), and \(900-2300=-1400\,\text{ft}\). 3. Their sum is \(900+1100-1200+900-1400=300\,\text{ft}\), which is the net change from start to finish. 4. Total climbing includes only the positive changes: \(900+1100+900=2900\,\text{ft}\). Descents reduce the net change but do not erase the climbing already completed.

Answer

a) Start: \(600\,\text{ft}\); A: \(1500\,\text{ft}\); B: \(2600\,\text{ft}\); C: \(1400\,\text{ft}\); D: \(2300\,\text{ft}\); Finish: \(900\,\text{ft}\) b) \(+900\,\text{ft}\), \(+1100\,\text{ft}\), \(-1200\,\text{ft}\), \(+900\,\text{ft}\), \(-1400\,\text{ft}\) c) \(+300\,\text{ft}\) d) The route includes \(2900\,\text{ft}\) of total ascent. The \(300\,\text{ft}\) value is only the net elevation change.
5351547
Find each missing starting number for the subtraction shown on the number line.
Figure for problem 535154

Hints

- Work backward from the ending value. - Undo a leftward jump by adding its length. - Check each answer by performing the subtraction from left to right.

Solution

1. In a), solve \(x-35=-10\). The starting number is \(-10+35=25\). 2. In b), solve \(x-60=-20\). The starting number is \(-20+60=40\). 3. In c), solve \(x-15=-45\). The starting number is \(-45+15=-30\).

Answer

a) \(25\) b) \(40\) c) \(-30\)
5351707
Complete each chain of jumps on the number line. Find every value marked with a question mark.
Figure for problem 535170

Hints

- Work one jump at a time. - Use a known middle value to work both forward and backward. - When reversing a jump, use the opposite operation.

Solution

1. In a), \(-50+120=70\), and \(70+(-90)=-20\). 2. In b), \(x+(-40)=10\), so \(x=50\). Then \(10+60=70\). 3. In c), the middle value satisfies \(y+(-70)=-60\), so \(y=10\). The starting value satisfies \(x+30=10\), so \(x=-20\).

Answer

a) \(70\) and \(-20\) b) \(50\) and \(70\) c) \(-20\) and \(10\)
5352557
Use the number line to find the missing starting value.
Figure for problem 535255

Hints

- Read the signed jump and endpoint from the diagram. - Undo the jump to recover the starting value.

Solution

1. The diagram shows an unknown start, a jump of \(-50\), and ending value \(-12\). 2. Solve \(x+(-50)=-12\), giving \(x=38\).

Answer

\(38\)
5352567
Use the number line to find the missing intermediate and ending values.
Figure for problem 535256

Hints

- Read the start and each signed jump directly from the diagram. - Find the first landing before applying the second jump.

Solution

1. The diagram starts at \(-15\) and first jumps \(+25\), landing at \(10\). 2. From \(10\), the next jump is \(-40\), landing at \(-30\).

Answer

Intermediate value: \(10\) Ending value: \(-30\)
5352597
The number-line diagram shows a temperature in the morning, its change by noon, and its change from noon to evening. What is the evening temperature?
Figure for problem 535259

Hints

- Read the starting temperature and both signed jumps from the diagram. - Follow the jumps in order from morning to noon to evening. - A positive jump raises the temperature and a negative jump lowers it.

Solution

1. The diagram shows a morning temperature of \(-8\,^\circ\text{F}\) and a rise of \(15\,^\circ\text{F}\). By noon, \(-8+15=7\), so the temperature is \(7\,^\circ\text{F}\). 2. The next jump is \(-4\,^\circ\text{F}\), so by evening \(7-4=3\). The evening temperature is \(3\,^\circ\text{F}\).

Answer

\(3\,^\circ\text{F}\)
5352767
The number line shows a temperature change from the previous evening to the morning low. Find the previous evening's temperature.
Figure for problem 535276

Hints

- Read the endpoint and signed jump from the number line. - Work backward from the morning low by reversing the displayed change. - Check that applying the displayed change to your answer reaches the shown endpoint.

Solution

1. Let the starting temperature be \(x\). The number line shows a fall of \(15\,^{\circ}\text{F}\) to \(-6\,^{\circ}\text{F}\), so \(x-15=-6\). 2. Add \(15\) to both sides: \(x=-6+15=9\). 3. The previous evening's temperature was \(9\,^{\circ}\text{F}\).

Answer

\(9\,^{\circ}\text{F}\)
5103577
Let \(A=\frac{5}{8}\) and \(B=\frac{6}{8}\). 1. Find the fraction \(M\) exactly halfway between \(A\) and \(B\). 2. Starting with \(M\), repeatedly find the midpoint between \(A\) and the most recently found midpoint. Explain why this produces infinitely many different fractions between \(\frac{5}{8}\) and \(\frac{6}{8}\).

Hints

- Find the mean of \(A\) and \(B\). - After finding one midpoint, use the interval between \(A\) and that midpoint. - Ask how the new midpoint compares with the preceding midpoint and with \(A\).

Solution

1. The midpoint is \(M=\frac{A+B}{2}=\frac{\frac{5}{8}+\frac{6}{8}}{2}=\frac{11}{16}\). 2. Check its position: \(\frac{5}{8}=\frac{10}{16}\) and \(\frac{6}{8}=\frac{12}{16}\), so \(\frac{10}{16}<\frac{11}{16}<\frac{12}{16}\). 3. At each later stage, use \(A\) and the newest midpoint as the two endpoints. Their midpoint is rational and lies strictly between them. Therefore, it is greater than \(A\), smaller than the previous midpoint, and different from every midpoint found earlier. This process can continue indefinitely.

Answer

1. \(M=\frac{11}{16}\) 2. Each new midpoint lies strictly between \(A\) and the preceding midpoint, so it is a new rational number inside the original interval. The process can be repeated indefinitely.
5106397
A student writes \(6 \frac{1}{4}-2 \frac{2}{3}=(6-2)+\left(\frac{2}{3}-\frac{1}{4}\right)=4+\left(\frac{8}{12}-\frac{3}{12}\right)=4 \frac{5}{12}\). Explain the error, then calculate the correct result.

Hints

- Compare the order of the fractional parts in the student's work with the original subtraction. - Subtraction is not commutative, so changing the order changes the value. - How can you regroup the first mixed number when its fractional part is too small to subtract?

Solution

1. The student reversed the order of the fractional parts. In the original subtraction, the fractional difference is \(\frac{1}{4}-\frac{2}{3}\), not \(\frac{2}{3}-\frac{1}{4}\). 2. Rewrite with denominator \(12\): \(6 \frac{3}{12}-2 \frac{8}{12}\). 3. Regroup one whole: \(5 \frac{15}{12}-2 \frac{8}{12}=3 \frac{7}{12}\).

Answer

The student incorrectly reversed \(\frac{1}{4}-\frac{2}{3}\) to \(\frac{2}{3}-\frac{1}{4}\). The correct result is \(3 \frac{7}{12}\).
5109237
Evaluate each expression. Pay attention to signs and simplify each result. a) \(2 \frac{1}{3}\cdot(4.5-6)\div\frac{7}{4}\) b) \(\frac{5}{12}\div\frac{25}{18}-0.4\cdot1 \frac{1}{2}\)

Hints

- Converting mixed numbers and decimals to fractions may make the arithmetic easier. - In a), determine the sign of the quantity in parentheses before multiplying. - After applying the order of operations, perform multiplication and division from left to right. - What sign results when a larger positive quantity is subtracted from a smaller one?

Solution

1. For a), \(2 \frac{1}{3}=\frac{7}{3}\) and \(4.5-6=-1.5=-\frac{3}{2}\). 2. Then \(\frac{7}{3}\cdot\left(-\frac{3}{2}\right)=-\frac{7}{2}\), and \(-\frac{7}{2}\div\frac{7}{4}=-\frac{7}{2}\cdot\frac{4}{7}=-2\). 3. For b), \(\frac{5}{12}\div\frac{25}{18}=\frac{5}{12}\cdot\frac{18}{25}=\frac{3}{10}\). 4. Also, \(0.4\cdot1 \frac{1}{2}=\frac{2}{5}\cdot\frac{3}{2}=\frac{3}{5}\). Therefore, \(\frac{3}{10}-\frac{3}{5}=-\frac{3}{10}\).

Answer

a) \(-2\) b) \(-\frac{3}{10}\), or \(-0.3\)
5121987
Divide the six number cards into two groups of three so that the sums of the groups are equal. \(1.2,\ \frac{3}{10},\ -1.5,\ 0.8,\ -0.2,\ -\frac{3}{5}\)

Hints

- Find the sum of all six numbers first. - Convert the fractions to decimals. - Look for three numbers that sum to half the total.

Solution

1. Write \(\frac{3}{10}=0.3\) and \(-\frac{3}{5}=-0.6\). 2. The sum of all six numbers is \(0\), so each group must have sum \(0\). 3. One group is \(1.2, 0.3, -1.5\), whose sum is \(0\). 4. The remaining group is \(0.8, -0.2, -0.6\), whose sum is also \(0\).

Answer

Group 1: \(1.2, \frac{3}{10}, -1.5\) Group 2: \(0.8, -0.2, -\frac{3}{5}\)
5178417
In \(m - s = d\), the minuend \(m\) is increased by \(10\). Determine how the subtrahend \(s\) must change so that the new difference has each property. a) The difference stays the same. b) The difference is \(15\) greater than before. c) The difference is \(5\) less than before.

Hints

- First determine the effect of increasing the minuend by \(10\). - Increasing the subtrahend lowers the difference; decreasing it raises the difference. - Work backward from the desired net change.

Solution

1. Increasing \(m\) by \(10\) first increases the difference by \(10\). To cancel that change, increase \(s\) by \(10\): \((m + 10) - (s + 10) = d\). 2. To make the total increase \(15\), the difference must increase by another \(5\). Decrease \(s\) by \(5\): \((m + 10) - (s - 5) = d + 15\). 3. To make the final difference \(5\) less than before, the change from the subtrahend must be \(-15\). Increase \(s\) by \(15\): \((m + 10) - (s + 15) = d - 5\).

Answer

a) Increase the subtrahend by \(10\). b) Decrease the subtrahend by \(5\). c) Increase the subtrahend by \(15\).
5181807
Decide whether each claim is true or false. Explain. a) If you add twice the opposite of a number to the original number, the result is the opposite of the original number. b) Two different integers can have the same opposite.

Hints

- Represent the original number with a variable. - Pair one copy of the number with one copy of its opposite. - Think of opposites as reflections across zero.

Solution

1. Let the number be \(n\). Then \(n+2(-n)=n-2n=-n\), so a) is true. 2. Every integer has exactly one opposite. If two integers had the same opposite, taking the opposite again would show that the integers are equal. Therefore, b) is false.

Answer

a) True; \(n+2(-n)=-n\). b) False; each integer has exactly one opposite.
5183097
Write one subtraction expression with a value of \(-35\) for each condition. a) Both numbers are positive integers. b) The starting number is a negative integer, and the number being subtracted is a positive integer. c) Both numbers are negative integers.

Hints

- For part a), the number being subtracted must be greater than the starting number. - Rewrite subtraction of a negative number as addition. - Choose one number first, and then determine what must be subtracted to reach \(-35\).

Solution

1. For a), the number being subtracted must be \(35\) greater than the starting number. One example is \(10-45=-35\). 2. For b), choose a negative starting number and a positive number whose distances from zero total \(35\). One example is \(-20-15=-35\). 3. For c), subtracting a negative adds its opposite. One example is \(-50-(-15)=-35\).

Answer

Possible answers are: a) \(10-45=-35\) b) \(-20-15=-35\) c) \(-50-(-15)=-35\)
5183207
Simon claims, “When I subtract one integer from another integer, the result is always less than the first integer.” Is Simon correct? Use at least two integer examples to justify your answer.

Hints

- Test a positive number, zero, and a negative number as the number being subtracted. - Subtracting a negative number is the same as adding its opposite. - One counterexample is enough to disprove an “always” statement, but provide two examples as requested.

Solution

1. Simon’s claim is not true for every integer being subtracted. 2. Subtracting a negative number makes the result greater. For example, \(5-(-3)=8\), and \(8>5\). 3. Subtracting zero leaves the first number unchanged. For example, \(5-0=5\). 4. The result is less than the first integer only when a positive integer is subtracted.

Answer

No. For example, \(5-(-3)=8\), which is greater than \(5\), and \(5-0=5\), which is equal to \(5\). The result is less than the first integer only when a positive integer is subtracted.
5183357
Consider the expressions: (I) \((-48)-(-122)\) (II) \((-210)-(+90)\) a) Rewrite each subtraction as addition and evaluate. b) What number must be subtracted from each answer in part a) to get \(10\)? c) For each answer in part a), what number must be subtracted to get its opposite?

Hints

- To subtract a number, add its opposite. - The opposite of a number has the same distance from zero and the opposite sign. - For parts b) and c), write an equation in the form \(\text{starting value}-x=\text{target value}\).

Solution

1. In a), (I) is \((-48)+122=74\), and (II) is \((-210)+(-90)=-300\). 2. In b), solve \(74-x=10\) to get \(x=64\). Solve \(-300-x=10\) to get \(x=-310\). 3. In c), the opposite of \(74\) is \(-74\). Solving \(74-x=-74\) gives \(x=148\). The opposite of \(-300\) is \(300\). Solving \(-300-x=300\) gives \(x=-600\).

Answer

a) (I) \(74\); (II) \(-300\) b) (I) \(64\); (II) \(-310\) c) (I) \(148\); (II) \(-600\)
5183367
Consider the expressions: (I) \((-15)-(+85)\) (II) \((+40)-(+110)\) a) Rewrite each subtraction as addition and evaluate. b) What number must be subtracted from the value of (I) to get the value of (II)? c) What number must be subtracted from the value of (II) to get the opposite of the value of (I)?

Hints

- Evaluate both expressions before answering parts b) and c). - Find the opposite of the value of Expression (I). - Represent the missing number in each subtraction with a variable.

Solution

1. In a), (I) is \((-15)+(-85)=-100\), and (II) is \(40+(-110)=-70\). 2. In b), solve \(-100-x=-70\). This gives \(x=-30\). 3. In c), the opposite of \(-100\) is \(100\). Solve \(-70-y=100\) to get \(y=-170\).

Answer

a) (I) \(-100\); (II) \(-70\) b) \(-30\) c) \(-170\)
5184297
Evaluate each expression one set of parentheses at a time. a) \(145-(34+156-20)-(55+45)\) b) \(-(-760+140)-[(24-110)+480]\)

Hints

- Evaluate the innermost parentheses first. - A negative sign outside parentheses changes the sign of the value inside. - Check the sign of every intermediate result.

Solution

1. In a), \(34+156-20=170\) and \(55+45=100\). Then \(145-170-100=-125\). 2. In b), \(-760+140=-620\), so \(-(-620)=620\). Also, \(24-110=-86\), and \(-86+480=394\). Therefore, \(620-394=226\).

Answer

a) \(-125\) b) \(226\)
5186397
Consider the expression \(2500 - [750 - (150 - 50)]\). First evaluate the expression. Then determine how its value changes if each of the four numbers is increased by \(10\). Explain why.

Hints

- Evaluate the expression from the innermost grouping outward. - Track how each inner difference changes before considering the outer subtraction. - Recall what happens when both numbers in a subtraction increase by the same amount.

Solution

1. Evaluate from the innermost grouping outward: \(150 - 50 = 100\), \(750 - 100 = 650\), and \(2500 - 650 = 1850\). 2. Increasing both numbers in \(150 - 50\) by \(10\) leaves that difference unchanged. 3. The value of \(750 - (150 - 50)\) increases by \(10\), because only its first number has a net increase. 4. In the outer subtraction, both the first number and the value being subtracted increase by \(10\). Therefore, the final difference remains \(1850\).

Answer

The original value is \(1850\), and it remains unchanged.
5187047
Write an expression with grouping symbols for the description, and then evaluate it. From the opposite of the sum of \(-315\) and \(105\), subtract the difference of \(45\) and \(-82\).

Hints

- The opposite of a number has the same absolute value and the opposite sign. - Use grouping symbols to separate the sum and the difference. - Evaluate the grouped expressions before the final subtraction.

Solution

1. The sum is \(-315+105=-210\), so its opposite is \(210\). 2. The difference is \(45-(-82)=127\). 3. The expression is \(-[(-315)+105]-[45-(-82)]\). 4. Evaluate: \(210-127=83\).

Answer

\(-[(-315)+105]-[45-(-82)]=83\)

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