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Difficulty: 1 – 5

5109797

Rewrite each expression in a convenient equivalent form, then evaluate it. a) \(17(-4) + 17(-6)\) b) \(-45 + 128 - 55\) c) \(0.5 \cdot 19 - 0.5 \cdot 9\)

Hints

- Look for a factor shared by two products. - Rewrite subtraction as addition of the opposite before rearranging addends. - Factoring can turn two products into one simpler product.

Solution

1. For a), factor out \(17\): \(17[(-4) + (-6)] = 17(-10) = -170\). 2. For b), rewrite subtraction as addition of the opposite, then rearrange and regroup: \((-45 - 55) + 128 = -100 + 128 = 28\). 3. For c), factor out \(0.5\): \(0.5(19 - 9) = 0.5 \cdot 10 = 5\).

Answer

a) \(-170\) b) \(28\) c) \(5\)

5112297

Consider the two expressions. Expression A: \(12\left(\frac{1}{4} - \frac{1}{3}\right)\) Expression B: \(12 \cdot \frac{1}{4} - 12 \cdot \frac{1}{3}\) a) Evaluate both expressions. b) Compare the results. Which property does this illustrate?

Hints

- For Expression A, evaluate the difference inside the parentheses first. - For Expression B, evaluate the two products first. - Which property relates multiplying a difference to subtracting two products?

Solution

1. For Expression A, subtract inside the parentheses: \(\frac{1}{4} - \frac{1}{3} = \frac{3}{12} - \frac{4}{12} = -\frac{1}{12}\). Then \(12\left(-\frac{1}{12}\right) = -1\). 2. For Expression B, evaluate each product: \(12 \cdot \frac{1}{4} = 3\) and \(12 \cdot \frac{1}{3} = 4\). Then \(3 - 4 = -1\). 3. Both expressions equal \(-1\). 4. The two equivalent forms illustrate the distributive property.

Answer

a) Expression A: \(-1\) Expression B: \(-1\) b) The results are equal. This illustrates the distributive property.

5113177

Lucas and Sophie are evaluating \(-5(1.2 - 4)\). Lucas says, “I can distribute. Then I get \(-5 \cdot 1.2 - 20 = -26\).” Sophie says, “The result should be positive.” 1. Determine who is correct. 2. Explain Lucas’s error in applying the distributive property.

Hints

- First evaluate the expression inside the parentheses as a check. - When distributing, multiply the outside factor by each term, including its sign. - Recall the sign of a product of two negative numbers.

Solution

1. Evaluate inside the parentheses: \(1.2 - 4 = -2.8\). Then \(-5(-2.8) = 14\), so Sophie is correct. 2. When distributing, the correct expression is \(-5 \cdot 1.2 - (-5 \cdot 4)\). 3. Therefore, \(-6 - (-20) = -6 + 20 = 14\). Lucas failed to account for the product of two negative factors.

Answer

1. Sophie is correct; the value is \(14\). 2. Lucas mishandled the signs. The second product contributes \(+20\), not \(-20\).

5113197

Consider the expressions \(A=-\frac{3}{4}(2.4-6.4)+1.5\) and \(B=-\frac{3}{4}\cdot2.4+\frac{3}{4}\cdot6.4+1.5\). 1. Explain without fully evaluating them why \(A\) and \(B\) must have the same value. 2. Evaluate \(A\). 3. Replace only the final \(1.5\) in \(B\) so that the value of the expression is \(0\). What number should replace it?

Hints

- Distribute the factor in \(A\) and compare the resulting terms with \(B\). - Pay close attention to the product of two negative numbers. - Find the value of the first two terms in \(B\), then determine its additive inverse.

Solution

1. Distributing \(-\frac{3}{4}\) in \(A\) gives \(-\frac{3}{4}\cdot2.4-\left(-\frac{3}{4}\cdot6.4\right)+1.5\), which simplifies to \(B\). 2. Evaluate \(A\): \(2.4-6.4=-4\), so \(-\frac{3}{4}(-4)+1.5=3+1.5=4.5\). 3. The part of \(B\) before the final number equals \(3\). Solve \(3+x=0\), giving \(x=-3\).

Answer

1. The distributive property shows that the expressions are equivalent. 2. \(A=4.5\) 3. Replace \(1.5\) with \(-3\).

5113307

Consider \((12.4 - 8.8) \div 4\). A student says, “It is easier to rewrite division by \(4\) as multiplication by \(\frac{1}{4}\), distribute, and divide each number inside the parentheses by \(4\) instead of subtracting first.” Evaluate the expression both ways. Which method requires less computation?

Hints

- Follow the grouping symbols when using the first method. - For the second method, view division by \(4\) as multiplication by \(\frac{1}{4}\). - Compare the number of arithmetic operations in the two methods.

Solution

1. Subtract first: \(12.4 - 8.8 = 3.6\), then \(3.6 \div 4 = 0.9\). 2. Distribute \(\frac{1}{4}\): \(12.4 \div 4 - 8.8 \div 4 = 3.1 - 2.2 = 0.9\). 3. Both methods are valid. Subtracting first requires one subtraction and one division, while distributing requires two divisions and one subtraction.

Answer

Both methods give \(0.9\). Subtracting first requires less computation.

5113377

Use the distributive property to evaluate each expression efficiently. a) Expand \(42\left(\frac{1}{7} + \frac{1}{6}\right)\). b) Factor \(17 \cdot 3.8 + 17 \cdot 6.2\).

Hints

- In a), multiply the outside factor by each addend. - In b), identify the common factor in both terms. - Choose the equivalent form that creates simpler arithmetic.

Solution

1. For a), distribute \(42\): \(42 \cdot \frac{1}{7} + 42 \cdot \frac{1}{6} = 6 + 7 = 13\). 2. For b), factor out \(17\): \(17(3.8 + 6.2) = 17 \cdot 10 = 170\).

Answer

a) \(13\) b) \(170\)

5113415

For each description, first write an expression using numbers, operation symbols, and parentheses. Then match your expression to every equivalent expression card. (A) The product of \(1.2\) and \(\frac{3}{4}\), increased by \(0.5\) (B) The sum of \(1.2\) and \(0.5\), multiplied by \(\frac{3}{4}\) (C) The quotient of \(1.2\) and the difference of \(\frac{3}{4}\) and \(0.5\) 1: \((1.2\times\frac{3}{4})+0.5\) 2: \((1.2+0.5)\times\frac{3}{4}\) 3: \(1.2\div(\frac{3}{4}-0.5)\) 4: \(\frac{3}{4}\times(0.5+1.2)\) 5: \(0.5+(1.2\times\frac{3}{4})\)

Hints

- Write each verbal description before looking for matching cards. - Pay attention to which quantities must be grouped before multiplication or division. - After writing each expression, look for cards that differ only by a valid reordering of addition or multiplication.

Solution

1. Description A translates to \((1.2\times\frac34)+0.5\). It matches Cards 1 and 5 because addition is commutative. 2. Description B translates to \((1.2+0.5)\times\frac34\). It matches Cards 2 and 4 because addition and multiplication are commutative. 3. Description C translates to \(1.2\div(\frac34-0.5)\). It matches Card 3.

Answer

(A) \((1.2\times\frac34)+0.5\); Cards 1 and 5 (B) \((1.2+0.5)\times\frac34\); Cards 2 and 4 (C) \(1.2\div(\frac34-0.5)\); Card 3

5113637

Use the distributive property to show that the two expressions are equivalent. Expression A: \(20\left(\frac{1}{4} - 0.2\right)\) Expression B: \(5 - 4\)

Hints

- Multiply \(20\) by each term inside the parentheses. - Evaluate \(20 \cdot \frac{1}{4}\) and \(20 \cdot 0.2\). - Compare the resulting expression with Expression B.

Solution

1. Distribute \(20\) in Expression A: \(20 \cdot \frac{1}{4} - 20 \cdot 0.2\). 2. Evaluate the first product: \(20 \cdot \frac{1}{4} = 5\). 3. Evaluate the second product: \(20 \cdot 0.2 = 4\). 4. The expanded form is \(5 - 4\), which is exactly Expression B. Both expressions equal \(1\).

Answer

Distributing \(20\) changes Expression A to \(5 - 4\), so the expressions are equivalent and both have value \(1\).

5114107

Without fully evaluating, decide whether each equation is true. Justify your decision using properties of operations. a) \(\frac{4}{9}\left(2.7 + \frac{9}{4}\right) = \frac{4}{9} \cdot 2.7 + 1\) b) \(15.3 - \left(4.2 + \frac{1}{3}\right) = (15.3 - 4.2) + \frac{1}{3}\)

Hints

- In a), distribute the factor to both terms inside the parentheses. - In b), determine how a subtraction sign affects every term in the grouped sum. - Look for reciprocal factors whose product is \(1\).

Solution

1. For a), distribute \(\frac{4}{9}\): \(\frac{4}{9} \cdot 2.7 + \frac{4}{9} \cdot \frac{9}{4}\). The second product is \(1\), so the equation is true. 2. For b), subtracting a sum means subtracting both addends: \(15.3 - 4.2 - \frac{1}{3}\). The right side adds \(\frac{1}{3}\), so the equation is false.

Answer

a) True; distributing gives \(\frac{4}{9} \cdot 2.7 + 1\). b) False; the correct form is \((15.3 - 4.2) - \frac{1}{3}\).

5114127

Let \(T = \frac{3}{4} \cdot 1.2 - \frac{3}{4} \cdot 0.4\). A student claims, “I can find \(T\) more easily by first subtracting \(0.4\) from \(1.2\), then multiplying the difference by \(0.75\).” Use a property of operations to determine whether the claim is correct. Would the same kind of rule work if the original expression used addition instead of subtraction? Explain.

Hints

- Identify the factor shared by both terms. - Use the distributive property in reverse. - Compare \(\frac{3}{4}\) with \(0.75\). - Recall whether the distributive property applies to sums as well as differences.

Solution

1. Both terms have the common factor \(\frac{3}{4}\). 2. Factor using the distributive property: \(\frac{3}{4} \cdot 1.2 - \frac{3}{4} \cdot 0.4 = \frac{3}{4}(1.2 - 0.4)\). 3. Since \(\frac{3}{4} = 0.75\), this is exactly the student’s method, so the claim is correct. 4. The distributive property also applies to addition: \(ab + ac = a(b + c)\). Therefore, the corresponding rule would also work with a plus sign.

Answer

The claim is correct because \(\frac{3}{4}\) can be factored: \(\frac{3}{4}(1.2 - 0.4)\). The same rule works for addition because \(ab + ac = a(b + c)\).

5116757

Use properties of operations to evaluate each expression efficiently. a) \(250 \cdot 13 \cdot 4\) b) \(0.8 \cdot 19 + 0.2 \cdot 19\) c) \(45 \cdot 11\)

Hints

- Look for factors that combine to make \(1000\). - In b), identify the common factor and add the decimal coefficients. - Rewrite \(11\) as a sum that makes multiplication easy.

Solution

1. For a), rearrange and regroup the factors: \((250 \cdot 4) \cdot 13 = 1000 \cdot 13 = 13{,}000\). 2. For b), factor out \(19\): \(19(0.8 + 0.2) = 19 \cdot 1 = 19\). 3. For c), write \(11 = 10 + 1\) and distribute: \(45(10 + 1) = 450 + 45 = 495\).

Answer

a) \(13{,}000\) b) \(19\) c) \(495\)

5119867

Use the distributive property to evaluate each expression efficiently. a) \(24\left(\frac{3}{8} + \frac{1}{6}\right)\) b) \(\frac{5}{12} \cdot \frac{4}{7} + \frac{5}{12} \cdot \frac{3}{7}\) c) \(15 \cdot \frac{7}{5} - 15 \cdot \frac{2}{3}\)

Hints

- In a), distributing avoids adding fractions with unlike denominators first. - In b), identify the common factor. - In c), multiply and simplify each product before subtracting.

Solution

1. For a), distribute \(24\): \(24 \cdot \frac{3}{8} + 24 \cdot \frac{1}{6} = 9 + 4 = 13\). 2. For b), factor out \(\frac{5}{12}\): \(\frac{5}{12}\left(\frac{4}{7} + \frac{3}{7}\right) = \frac{5}{12} \cdot 1 = \frac{5}{12}\). 3. For c), factor out \(15\): \(15\left(\frac{7}{5} - \frac{2}{3}\right)\). Distributing directly gives \(21 - 10 = 11\).

Answer

a) \(13\) b) \(\frac{5}{12}\) c) \(11\)

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

\(4\)

5122507

Evaluate \(8(12.5 + 5)\) in two ways and compare the results. a) Add inside the parentheses first, then multiply. b) Apply the distributive property first.

Hints

- For a), evaluate the sum inside the parentheses first. - For b), multiply \(8\) by each addend. - Equivalent methods must produce the same value.

Solution

1. Add first: \(8(12.5 + 5) = 8(17.5) = 140\). 2. Distribute first: \(8 \cdot 12.5 + 8 \cdot 5 = 100 + 40 = 140\). 3. Both methods give the same value, as expected for equivalent expressions.

Answer

a) \(140\) b) \(140\) The results are equal.

5122577

Use the distributive property to evaluate each expression mentally. a) \(12 \cdot 101\) b) \(\frac{5}{9} \cdot 14 - \frac{5}{9} \cdot 5\)

Hints

- Rewrite \(101\) using \(100\) and \(1\). - In b), identify the common factor. - Choose the equivalent form that creates the simplest arithmetic.

Solution

1. For a), write \(101 = 100 + 1\): \(12(100 + 1) = 1200 + 12 = 1212\). 2. For b), factor out \(\frac{5}{9}\): \(\frac{5}{9}(14 - 5) = \frac{5}{9} \cdot 9 = 5\).

Answer

a) \(1212\) b) \(5\)

5122597

Evaluate each expression in two ways: first by calculating the products separately, and then by factoring out the common factor. a) \(14 \cdot 11 + 14 \cdot 9\) b) \(6.5 \cdot 8 - 6.5 \cdot 6\) c) \((-1.5) \cdot 4 + (-1.5) \cdot 6\)

Hints

- For the direct method, evaluate multiplication before addition or subtraction. - For the factoring method, identify the factor shared by both terms. - Compare the two methods after completing each part.

Solution

1. For a), direct calculation gives \(154 + 126 = 280\). Factoring gives \(14(11 + 9) = 14 \cdot 20 = 280\). 2. For b), direct calculation gives \(52 - 39 = 13\). Factoring gives \(6.5(8 - 6) = 6.5 \cdot 2 = 13\). 3. For c), direct calculation gives \(-6 + (-9) = -15\). Factoring gives \((-1.5)(4 + 6) = -1.5 \cdot 10 = -15\).

Answer

a) \(280\) b) \(13\) c) \(-15\)

5122685

Write an expression for each description, then evaluate it efficiently. a) Find the product of \(16\) and \(27\), and subtract the product of \(16\) and \(17\). b) Add the product of \(0.2\) and \(35\) to the product of \(0.2\) and \(15\).

Hints

- Translate each description into one expression before evaluating. - Keep each stated product grouped as a product in your written expression. - After the expression is written, look for a common factor that can make the arithmetic shorter.

Solution

1. For a), write \(16\times27-16\times17\). Factoring the common factor gives \(16(27-17)=16\times10=160\). 2. For b), write \(0.2\times35+0.2\times15\). Factoring gives \(0.2(35+15)=0.2\times50=10\).

Answer

a) Expression: \(16\times27-16\times17\); value: \(160\) b) Expression: \(0.2\times35+0.2\times15\); value: \(10\)

5124667

Determine whether each pair of expressions is equivalent. Show your reasoning by simplifying both expressions. a) \(4(a+2)+3a\) and \(7a+8\) b) \(15y-6y+2\) and \(3(3y+2)-4\)

Hints

- Use the distributive property to remove parentheses. - Combine like terms in each expression. - Equivalent expressions simplify to the same result.

Solution

1. For part a, simplify the first expression: \(4(a+2)+3a=4a+8+3a=7a+8\). It matches the second expression, so the pair is equivalent. 2. For part b, simplify the first expression: \(15y-6y+2=9y+2\). 3. Simplify the second expression: \(3(3y+2)-4=9y+6-4=9y+2\). 4. Both expressions in part b simplify to \(9y+2\), so the pair is equivalent.

Answer

a) Equivalent; both simplify to \(7a+8\). b) Equivalent; both simplify to \(9y+2\).

5124727

Simplify each expression. For every step, name the property or operation you used: the distributive property, the commutative property, or combining like terms. a) \(4(2x - 5) + 3x\) b) \(15y - 3(5y - 2)\) c) \(8z + 4 - 2(4z + 2)\)

Hints

- Pay close attention to the sign before each set of parentheses. - Combine only terms with the same variable part. - When a negative factor is distributed, how does it affect each term inside the parentheses?

Solution

1. For a), use the distributive property: \(4(2x - 5) + 3x = 8x - 20 + 3x\). Use the commutative property of addition to reorder the terms: \(8x - 20 + 3x = 8x + 3x - 20\). Combine like terms: \(8x + 3x - 20 = 11x - 20\). 2. For b), use the distributive property: \(15y - 3(5y - 2) = 15y - 15y + 6\). Combine like terms: \(15y - 15y + 6 = 6\). 3. For c), use the distributive property: \(8z + 4 - 2(4z + 2) = 8z + 4 - 8z - 4\). Use the commutative property of addition to reorder the terms: \(8z + 4 - 8z - 4 = 8z - 8z + 4 - 4\). Combine like terms: \(8z - 8z + 4 - 4 = 0\).

Answer

a) \(11x - 20\) b) \(6\) c) \(0\)

5124787

Which expressions are equivalent to \(4x+10\)? Simplify each expression to justify your choices. a) \(2x+10+2x\) b) \(2(2x+5)\) c) \(15x-11x+12-2\) d) \(4(x+10)\) e) \(x+x+x+x+7+3\)

Hints

- Combine variable terms and constant terms separately. - Use the distributive property for expressions with parentheses. - Compare each simplified expression with \(4x+10\).

Solution

1. For part a, \(2x+10+2x=4x+10\), so it is equivalent. 2. For part b, \(2(2x+5)=4x+10\), so it is equivalent. 3. For part c, \(15x-11x+12-2=4x+10\), so it is equivalent. 4. For part d, \(4(x+10)=4x+40\), so it is not equivalent. 5. For part e, \(x+x+x+x+7+3=4x+10\), so it is equivalent.

Answer

The equivalent expressions are a), b), c), and e). Expression d) simplifies to \(4x+40\), so it is not equivalent.

5124867

Two expressions are equivalent if they have the same value for every value of the variable. Determine by simplifying whether \(T_1 = 4(k + 2) - 3k - 8\) and \(T_2 = k\) are equivalent. Then find the value of \(T_1\) when \(k = 12.5\).

Hints

- If one expression simplifies to exactly the other expression, what does that show? - For the evaluation, can you use the simpler equivalent expression? - Look for terms that cancel.

Solution

1. Expand and combine like terms: \(T_1 = 4k + 8 - 3k - 8 = k\). 2. Since \(T_1\) simplifies to \(T_2\), the expressions are equivalent. 3. When \(k = 12.5\), \(T_1 = k = 12.5\).

Answer

The expressions are equivalent because \(T_1\) simplifies to \(k\). For \(k = 12.5\), the value of \(T_1\) is \(12.5\).

5126047

Which expressions have the same value as \(6x-12\) for every value of \(x\)? Select all that apply. a) \(3(2x-4)\) b) \(6(x-2)\) c) \(10x-12-4x\) d) \(2x+4x-6\cdot2\) e) \(6x-12\) f) \(x(6-12)\) g) \(12x\div2-12\)

Hints

- Simplify each expression completely. - Use the distributive property for expressions with parentheses. - Combine variable terms and constant terms separately. - Apply multiplication and division before addition and subtraction.

Solution

1. \(3(2x-4)=6x-12\), so a) is equivalent. 2. \(6(x-2)=6x-12\), so b) is equivalent. 3. \(10x-12-4x=6x-12\), so c) is equivalent. 4. \(2x+4x-6\cdot2=6x-12\), so d) is equivalent. 5. Expression e) is identical to the given expression. 6. \(x(6-12)=-6x\), so f) is not equivalent. 7. \(12x\div2-12=6x-12\), so g) is equivalent.

Answer

a), b), c), d), e), and g)

5126317

Try this number trick: 1. Choose any number \(x\). 2. Add \(5\). 3. Multiply the result by \(3\). 4. Subtract \(15\). 5. Subtract twice your original number \(x\). a) Try the trick with two different starting numbers. What do you notice? b) Write an expression for the result after each step. Simplify the final expression to explain why the trick works.

Hints

- Translate each direction into an algebraic expression. - Use the distributive property when you multiply the sum by \(3\). - Make sure the multiplication in step \(3\) applies to the entire previous result.

Solution

1. For \(x=2\): \(2+5=7\), \(7\cdot3=21\), \(21-15=6\), and \(6-2\cdot2=2\). 2. For \(x=-4\): \(-4+5=1\), \(1\cdot3=3\), \(3-15=-12\), and \(-12-2(-4)=-4\). 3. In general, the expressions after the five steps are \(x\), \(x+5\), \(3(x+5)=3x+15\), \(3x\), and \(3x-2x=x\). 4. The final result is always the original number.

Answer

a) For example, starting with \(2\) gives \(2\), and starting with \(-4\) gives \(-4\). The trick returns the starting number. b) The expressions are \(x\), \(x+5\), \(3x+15\), \(3x\), and \(x\).

5129017

Determine whether each statement is correct. For each incorrect statement, describe the error and give the correct result. a) \(7x-x=7\) b) \(\frac{x}{3}+\frac{x}{3}=\frac{2x}{3}\) c) \(3(2a+4)=6a+4\) d) \(4y+3y=7y\)

Hints

- Write an implied coefficient of \(1\) when needed. - For fractions with the same denominator, add the numerators. - Apply the distributive property to every term inside parentheses.

Solution

1. Part a is incorrect. Since \(x=1x\), \(7x-x=(7-1)x=6x\). The variable should not disappear. 2. Part b is correct. The fractions have the same denominator, so their numerators add: \(\frac{x+x}{3}=\frac{2x}{3}\). 3. Part c is incorrect. The factor \(3\) must multiply both terms: \(3(2a+4)=6a+12\). 4. Part d is correct because \(4y+3y=(4+3)y=7y\).

Answer

a) Incorrect; \(7x-x=6x\). b) Correct. c) Incorrect; \(3(2a+4)=6a+12\). d) Correct.

5152207

Simplify each expression completely. a) \(12x-5y+(3x-4y)\) b) \(8a-(2a-7b)-10b\) c) \(-15m+4n-(-6m+4n)\)

Hints

- Check the sign before each set of parentheses. - A negative sign before parentheses reverses every sign inside. - Combine only terms with identical variable parts.

Solution

1. For part a, remove the parentheses and combine like terms: \(12x-5y+3x-4y=15x-9y\). 2. For part b, distribute the negative sign and combine like terms: \(8a-2a+7b-10b=6a-3b\). 3. For part c, distribute the negative sign and combine like terms: \(-15m+4n+6m-4n=-9m\).

Answer

a) \(15x-9y\) b) \(6a-3b\) c) \(-9m\)

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 \(16-(-8)\).

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

\(-70\)

5223937

Rewrite each expression as a sum or difference so that no coefficient has an absolute value greater than \(1\). a) \(5x\) b) \(3a^2+2b\) c) \(4m-3n\) d) \(2(p+q)\)

Hints

- A coefficient tells how many copies of a variable expression are being added. - Think of each power, such as \(a^2\), as one complete unit. - For a grouped expression, repeat the entire group.

Solution

1. Interpret multiplication as repeated addition: \(5x=x+x+x+x+x\). 2. Write three copies of \(a^2\) and two copies of \(b\): \(3a^2+2b=a^2+a^2+a^2+b+b\). 3. Write four copies of \(m\) and subtract three copies of \(n\): \(4m-3n=m+m+m+m-n-n-n\). 4. Treat \(p+q\) as one grouped expression and add two copies: \(2(p+q)=(p+q)+(p+q)\).

Answer

a) \(x+x+x+x+x\) b) \(a^2+a^2+a^2+b+b\) c) \(m+m+m+m-n-n-n\) d) \((p+q)+(p+q)\)

5225397

A wire-frame square pyramid has a square base with side length \(a\) and four lateral edges of length \(s\). 1. Write an expression for the total wire length \(L\). 2. Simplify the expression when each lateral edge is twice as long as a base edge.

Hints

- Count the edges in the square base and the edges from the base to the apex. - A square pyramid has eight edges in all. - Translate “twice as long” into an equation relating \(s\) and \(a\). - Combine like terms after substitution.

Solution

1. The square base has four edges of length \(a\), and the pyramid has four lateral edges of length \(s\). Thus, \(L=4a+4s=4(a+s)\). 2. If \(s=2a\), then \(L=4a+4(2a)=4a+8a=12a\).

Answer

1. \(L=4a+4s\), or \(L=4(a+s)\) 2. \(L=12a\)

5226807

Rewrite each expression in the requested equivalent form. a) Write \(p - q - r - 10\) as a sum of four terms. b) Write \(15 + a + b\) using subtraction signs between all terms. c) Write \(x - (-y) - z\) using only addition between terms.

Hints

- Subtraction can be rewritten as addition of the opposite. - Addition can be rewritten as subtraction of the opposite. - Simplify a double negative before completing the requested form.

Solution

1. For a), rewrite each subtraction as addition of the opposite: \(p + (-q) + (-r) + (-10)\). 2. For b), adding a value is equivalent to subtracting its opposite: \(15 - (-a) - (-b)\). 3. For c), subtracting \(-y\) is adding \(y\), and subtracting \(z\) is adding \(-z\): \(x + y + (-z)\).

Answer

a) \(p + (-q) + (-r) + (-10)\) b) \(15 - (-a) - (-b)\) c) \(x + y + (-z)\)

5229107

For values of \(x\) and \(y\) that make all lengths positive and form a valid triangle, a student claims that these figures have equal perimeters: A square with side length \(s=1.5x+2y\) A triangle with side lengths \(2x+3y\), \(3x+4y\), and \(x+y\) Check the claim by writing and simplifying an expression for each perimeter.

Hints

- Multiply a square’s side length by \(4\). - Add all three side lengths of the triangle. - Compare the simplified expressions.

Solution

1. The square’s perimeter is \(4(1.5x+2y)=6x+8y\). 2. The triangle’s perimeter is \((2x+3y)+(3x+4y)+(x+y)\). 3. Combine like terms for the triangle: \(2x+3x+x=6x\) and \(3y+4y+y=8y\). 4. Both perimeters equal \(6x+8y\), so the claim is true.

Answer

The claim is true. Both perimeters are \(6x+8y\).

5229117

For values of \(x\) and \(y\) that make the lengths positive and form a valid triangle, the side lengths are \(5x+2y\), \(3x-4y\), and \(2x+3y\). Write and simplify an expression for the perimeter \(P\).

Hints

- A triangle’s perimeter is the sum of its side lengths. - Combine all \(x\)-terms and all \(y\)-terms separately. - Keep the negative sign on \(-4y\).

Solution

1. Add the three side lengths: \(P=(5x+2y)+(3x-4y)+(2x+3y)\). 2. Combine the \(x\)-terms: \(5x+3x+2x=10x\). 3. Combine the \(y\)-terms: \(2y-4y+3y=y\). 4. Therefore, \(P=10x+y\).

Answer

\(P=10x+y\)

5229157

For values of \(x\) and \(y\) that make the lengths positive and form a valid quadrilateral, its side lengths are \(s_1=3x+2\) \(s_2=2y-1\) \(s_3=x+y+4\) \(s_4=2x-y+3\). Write and simplify an expression for the perimeter \(P\).

Hints

- Perimeter is the sum of all side lengths. - Combine the \(x\)-terms, \(y\)-terms, and constants separately. - Keep each term’s sign.

Solution

1. Add all four side lengths: \(P=(3x+2)+(2y-1)+(x+y+4)+(2x-y+3)\). 2. Combine the \(x\)-terms: \(3x+x+2x=6x\). 3. Combine the \(y\)-terms: \(2y+y-y=2y\). 4. Combine the constants: \(2-1+4+3=8\). 5. Therefore, \(P=6x+2y+8\).

Answer

\(P=6x+2y+8\)

5229457

Simplify each expression and identify which expressions are equivalent. A: \(-15a-(-6a)\) B: \(-4a+(-5a)\) C: \(-(12a-3a)\) D: \(2a-11a\) E: \(-a-(+10a)\)

Hints

- Simplify each expression separately. - Subtracting a negative term changes to addition. - Compare the simplified results.

Solution

1. \(A=-15a+6a=-9a\). 2. \(B=-4a-5a=-9a\). 3. \(C=-12a+3a=-9a\). 4. \(D=2a-11a=-9a\). 5. \(E=-a-10a=-11a\). 6. Therefore, A, B, C, and D are equivalent. E is not equivalent to them.

Answer

A, B, C, and D are equivalent because each simplifies to \(-9a\). Expression E simplifies to \(-11a\).

5229737

Simplify the expression. \(2x-[5x-(3x+4)]\)

Hints

- Work from the innermost grouping symbols outward. - A negative sign before parentheses reverses every sign inside. - Simplify inside the brackets before removing them.

Solution

1. Work from the innermost parentheses: \(2x-[5x-3x-4]\). 2. Combine like terms inside the brackets: \(2x-[2x-4]\). 3. Distribute the negative sign: \(2x-2x+4\). 4. The variable terms cancel, so the result is \(4\).

Answer

\(4\)

5230289

Simplify each expression. a) \((-3x)\cdot(-2y)-8xy\) b) \(4a\cdot(-b)+(-2a)\cdot(-3b)\) c) \((-1)\cdot m\cdot n\cdot(-1)-mn\)

Hints

- Multiply before adding or subtracting. - Simplify each product first. - Only terms with exactly the same variable part can be combined. - Track the sign of each product carefully.

Solution

1. For part a, multiply first: \((-3x)\cdot(-2y)=6xy\). Then \(6xy-8xy=-2xy\). 2. For part b, evaluate each product: \(4a\cdot(-b)=-4ab\) and \((-2a)\cdot(-3b)=6ab\). Then \(-4ab+6ab=2ab\). 3. For part c, the two factors of \(-1\) have product \(1\), so the first term is \(mn\). Therefore, \(mn-mn=0\).

Answer

a) \(-2xy\) b) \(2ab\) c) \(0\)

5234286

For each item, first identify the two structural parts of the quotient, then rewrite the same quotient in the other notation. a) In \(18\div x\), identify the dividend and divisor, then rewrite it as a fraction. b) In \(\frac{a+5}{3}\), identify the numerator and denominator, then rewrite it using division notation. c) In \(\frac{2m-1}{n+4}\), identify the numerator and denominator, then rewrite it using division notation.

Hints

- A quotient has two structural parts: what is being divided and what it is divided by. - In fraction notation, those parts are the numerator and denominator. - Preserve grouped expressions such as \(a+5\) or \(n+4\) as single parts when changing notation.

Solution

1. In \(18\div x\), the dividend is \(18\) and the divisor is \(x\). The same quotient is \(\frac{18}{x}\). 2. In \(\frac{a+5}{3}\), the numerator is \(a+5\) and the denominator is \(3\). The same quotient is \((a+5)\div3\). 3. In \(\frac{2m-1}{n+4}\), the numerator is \(2m-1\) and the denominator is \(n+4\). The same quotient is \((2m-1)\div(n+4)\).

Answer

a) Dividend: \(18\); divisor: \(x\); \(\frac{18}{x}\) b) Numerator: \(a+5\); denominator: \(3\); \((a+5)\div3\) c) Numerator: \(2m-1\); denominator: \(n+4\); \((2m-1)\div(n+4)\)

5240707

A streaming service offers two monthly plans. Basic: a monthly fee of \(\$G\) plus \(\$k\) for each rented movie. Premium: a flat monthly fee of \(\$P\) with no additional charge per movie. Let \(K_P\) be the Premium plan's monthly cost. a) Write an expression for the Basic plan's monthly cost \(K_B\) when a customer rents \(n\) movies. b) Write an expression for the difference \(K_P-K_B\). c) What does it mean when the expression from part b) equals \(0\)?

Hints

- Add the fixed fee and the per-movie charges for the Basic plan. - Subtract the entire Basic expression from the Premium cost. - A difference of \(0\) means the two quantities are equal.

Solution

1. The Basic plan costs \(K_B=G+nk\). 2. The Premium plan costs \(K_P=P\). 3. The difference is \(K_P-K_B=P-(G+nk)=P-G-nk\). 4. If the difference is \(0\), then \(P=G+nk\), so the plans cost the same for that number of movies.

Answer

a) \(K_B=G+nk\) b) \(P-(G+nk)=P-G-nk\) c) Both plans have the same monthly cost for that number of movies.

5241427

Try this number puzzle: Choose a number, multiply it by \(3\), add \(15\), subtract the original number, and divide the result by \(2\). a) What result do you get when the starting number is \(4\)? b) Write and simplify an expression for the puzzle. c) Use the simplified expression to describe how the final result depends on the starting number.

Hints

- First follow the directions with the number from part a). - Use \(x\) to represent an unknown starting number. - Combine the terms containing \(x\) before dividing. - Interpret what an expression of the form \(x+\text{constant}\) means.

Solution

1. Starting with \(4\) gives \(\frac{3\cdot4+15-4}{2}=\frac{23}{2}=11.5\). 2. For a starting number \(x\), the expression is \(\frac{3x+15-x}{2}\). 3. Combine like terms in the numerator: \(\frac{2x+15}{2}=x+7.5\). 4. The final result is always \(7.5\) greater than the starting number.

Answer

a) \(11.5\) b) \(\frac{3x+15-x}{2}=x+7.5\) c) The final result is always \(7.5\) greater than the starting number.

5241457

Choose a number. Multiply it by \(4\), add \(12\), multiply that sum by \(5\), and subtract \(60\). Once you report the final result, someone can immediately determine your starting number. Explain how to recover the starting number, and justify your rule with an algebraic expression.

Hints

- Represent the starting number with \(x\). - Translate the operations in order into one expression. - Distribute and combine like terms. - What factor remains in front of \(x\)?

Solution

1. Let the starting number be \(x\). The final result is \(5(4x+12)-60\). 2. Distribute: \(5(4x+12)-60=20x+60-60\). 3. Simplify to \(20x\). 4. Since the final result is \(20\) times the starting number, divide the final result by \(20\) to recover \(x\).

Answer

Divide the final result by \(20\). The expression \(5(4x+12)-60\) simplifies to \(20x\).

5279597

Change the sign before each set of parentheses to the opposite sign without changing the value of the expression. 1) \(12-(4a-5b)\) 2) \(x+(2y-3z+1)\) 3) \(3k-(k-m)\)

Hints

- Think of a negative sign before parentheses as multiplication by \(-1\). - Reverse every sign inside when you reverse the sign outside. - Expand your result to check that it matches the original expression.

Solution

1. Change subtraction to addition and reverse every sign inside the parentheses: \(12+(-4a+5b)\). 2. Change addition to subtraction and reverse every sign inside the parentheses: \(x-(-2y+3z-1)\). 3. Change subtraction to addition and reverse every sign inside the parentheses: \(3k+(-k+m)\).

Answer

1) \(12+(-4a+5b)\) 2) \(x-(-2y+3z-1)\) 3) \(3k+(-k+m)\)

5317467

Luisa is building \(4\) identical wooden picture frames. Each frame needs \(2\) long strips and \(2\) short strips. The two calculation trees show the strip lengths, in meters, and two different methods. a) Explain what each step in tree a) and tree b) represents in the situation. b) Find every missing value in both trees and determine the total length of wood needed. c) Name the property that guarantees both methods give the same result. Write the expression represented by each tree.

Figure for problem 531746

Hints

- Determine how many long and short strips are needed for all four frames. - Work from the leaves toward the root of each tree. - One tree calculates one frame first; the other calculates each strip type for all frames. - Which property distributes multiplication across a sum?

Solution

1. Tree a) finds the wood needed for one frame: \(2\cdot1.2=2.4\) meters of long strips and \(2\cdot0.85=1.7\) meters of short strips. It adds these to get \(4.1\) meters per frame, then multiplies by \(4\). 2. Tree a) gives \(4(2.4+1.7)=4\cdot4.1=16.4\,\text{m}\). 3. Tree b) uses \(8\) long strips and \(8\) short strips for all four frames. The partial lengths are \(8\cdot1.2=9.6\,\text{m}\) and \(8\cdot0.85=6.8\,\text{m}\). 4. Tree b) gives \(9.6+6.8=16.4\,\text{m}\). 5. The distributive property shows the methods are equivalent. The expressions are \(4(2\cdot1.2+2\cdot0.85)\) and \(8\cdot1.2+8\cdot0.85\).

Answer

a) Tree a) finds the wood for one frame and then multiplies by \(4\). Tree b) finds the total lengths of all long and all short strips separately, then adds them. b) Tree a) has missing values \(2.4\), \(1.7\), \(4.1\), and \(16.4\). Tree b) has missing values \(9.6\), \(6.8\), and \(16.4\). The total length is \(16.4\,\text{m}\). c) Distributive property; \(4(2\cdot1.2+2\cdot0.85)\) and \(8\cdot1.2+8\cdot0.85\).

5101257

A city has a population of \(36{,}000\). During the first year, the population grows by \(0.5\%\). a) What is the population after the first year? b) During the second year, the new population grows by \(p\%\). Write an expression for the population after the second year.

Hints

- Treat the two yearly changes separately. - The second-year increase is based on the population after the first year. - Write the percent increase using the variable \(p\). - Leave the variable in the final expression.

Solution

1. For part a, the first-year increase is \(36{,}000 \cdot 0.005 = 180\). The population after the first year is \(36{,}000 + 180 = 36{,}180\). 2. For part b, the second-year increase is \(36{,}180 \cdot \frac{p}{100} = 361.8p\). 3. Therefore, the population after the second year is \(36{,}180 + 361.8p\), or equivalently \(36{,}180\left(1 + \frac{p}{100}\right)\).

Answer

a) The population after the first year is \(36{,}180\). b) The population after the second year is \(36{,}180\left(1 + \frac{p}{100}\right)\).

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) \(-15-30+45\)

5107017

Rewrite each expression in a more efficient equivalent form, then evaluate it. a) \(16 \cdot 12 + 16 \cdot 8\) b) \(25 \cdot 37 \cdot 4\) c) \(45 \cdot 99\)

Hints

- In a), look for a factor shared by both products. - In b), which two factors multiply to make \(100\)? - In c), rewrite \(99\) using \(100\) and \(1\).

Solution

1. For a), factor out the common factor \(16\): \(16(12 + 8) = 16 \cdot 20 = 320\). 2. For b), use the commutative and associative properties to group \(25\) and \(4\): \((25 \cdot 4) \cdot 37 = 100 \cdot 37 = 3700\). 3. For c), rewrite \(99\) as \(100 - 1\) and distribute: \(45(100 - 1) = 45 \cdot 100 - 45 \cdot 1 = 4500 - 45 = 4455\).

Answer

a) \(320\) b) \(3700\) c) \(4455\)

5113217

Use the distributive property to evaluate the expression efficiently. \(12\left(\frac{3}{4} - 0.5 + \frac{1}{6}\right)\)

Hints

- Multiply the outside factor by every term inside the parentheses. - Keep each operation sign from inside the parentheses. - Each product simplifies to a whole number.

Solution

1. Distribute \(12\) to each term: \(12 \cdot \frac{3}{4} - 12 \cdot 0.5 + 12 \cdot \frac{1}{6}\). 2. Evaluate the products: \(12 \cdot \frac{3}{4} = 9\), \(12 \cdot 0.5 = 6\), and \(12 \cdot \frac{1}{6} = 2\). 3. Combine the results: \(9 - 6 + 2 = 5\).

Answer

\(5\)

5113617

Analyze the structure of the expression and identify whether it is a sum, difference, product, or quotient at its highest level. Then evaluate efficiently by looking for a pattern. \(\left[12.5 - \left(-\frac{3}{7}\right) + 7.5\right] + \left[2.5 - \left(-\frac{3}{7}\right) - 22.5\right] - \left(-\frac{3}{7}\right)\)

Hints

- Notice the expression that appears repeatedly. - Combine the decimal values before evaluating the repeated fraction. - Temporarily replace the repeated expression with a variable.

Solution

1. The expression is a difference at its highest level because the final operation subtracts \(-\frac{3}{7}\). 2. Let \(a = -\frac{3}{7}\). The expression becomes \([12.5-a+7.5]+[2.5-a-22.5]-a\). 3. Combine the decimal terms: \((20-a)+(-20-a)-a\). 4. Simplify: \(20-20-a-a-a=-3a\). 5. Substitute \(a=-\frac{3}{7}\): \(-3\left(-\frac{3}{7}\right)=\frac{9}{7}=1\frac{2}{7}\).

Answer

It is a difference at its highest level. Its value is \(\frac{9}{7}\), or \(1\frac{2}{7}\).

5113627

Without calculating the final value, explain why the two expressions must be equal. Expression A: \(\left(-\frac{3}{4}+0.5\right)\cdot(-2)\) Expression B: \(\left|-\frac{3}{4}+0.5\right|\cdot2\)

Hints

- First determine the sign of the quantity inside the parentheses. - What does multiplying by \(-2\) do to a negative value? - What does absolute value do to a negative value? - Compare the resulting forms rather than calculating the final number.

Solution

1. Let \(x=-\frac{3}{4}+0.5\). Since \(0.5=\frac{1}{2}<\frac{3}{4}\), \(x<0\). 2. Expression A is \(x\cdot(-2)=(-x)\cdot2\). 3. Because \(x<0\), \(|x|=-x\). Therefore, Expression B is also \((-x)\cdot2\). 4. Both expressions have the same form, so they must have the same value.

Answer

Let \(x=-\frac{3}{4}+0.5\). Since \(x<0\), \(|x|=-x\). Both expressions therefore equal \((-x)\cdot2\), so they have the same value.

5113647

Determine whether expressions A and B have the same value. Justify your answer using sign rules for powers and products. Expression A: \((-1)^3\cdot4.2\) Expression B: \(4.2\cdot(-1)^7\)

Hints

- What happens when \(-1\) is raised to an odd power? - Does an even or odd exponent change the sign? - Does changing the order of factors change a product?

Solution

1. Since \(3\) is odd, \((-1)^3=-1\). Therefore, Expression A is \(-1\cdot4.2=-4.2\). 2. Since \(7\) is odd, \((-1)^7=-1\). Therefore, Expression B is \(4.2\cdot(-1)=-4.2\). 3. Both expressions have the same value.

Answer

Yes. Both expressions equal \(-4.2\) because both odd powers of \(-1\) equal \(-1\), and multiplication can be written in either factor order.

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 \(3\).

5114117

Determine which equations are true. Justify each decision without fully evaluating the final value. a) \(0.8 \cdot \frac{3}{7} \cdot 1.25 = \frac{3}{7}\) b) \(\frac{5}{6} \div \left(\frac{1}{3} \cdot 2\right) = \frac{5}{6} \div \frac{1}{3} \cdot 2\)

Hints

- In a), rearrange the factors to pair the two decimals. - Convert the decimals to fractions to check whether their product is \(1\). - In b), compare dividing by twice a number with dividing by the number and then doubling.

Solution

1. For a), rearrange and regroup the factors: \((0.8 \cdot 1.25)\frac{3}{7}\). Since \(0.8 = \frac{4}{5}\) and \(1.25 = \frac{5}{4}\), their product is \(1\). The equation is true. 2. For b), dividing by a product is equivalent to dividing successively by both factors: \(a \div (bc) = a \div b \div c\). The right side divides by \(\frac{1}{3}\) and then multiplies by \(2\), so the equation is false.

Answer

a) True, because \(0.8 \cdot 1.25 = 1\). b) False, because dividing by \(\frac{1}{3} \cdot 2\) is not the same as dividing by \(\frac{1}{3}\) and then multiplying by \(2\).

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 \(\frac{3}{8}\).

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) \(2.5\)

5122517

Evaluate \(48\left(\frac{7}{12} - \frac{5}{16}\right)\) in two ways. a) Subtract the fractions using a common denominator, then multiply by \(48\). b) Distribute \(48\) first, then subtract.

Hints

- Find the least common denominator of \(12\) and \(16\). - When distributing, simplify each product before subtracting. - Compare how much fraction arithmetic each method requires.

Solution

1. For a), the least common denominator is \(48\): \(\frac{7}{12} - \frac{5}{16} = \frac{28}{48} - \frac{15}{48} = \frac{13}{48}\). Then \(48 \cdot \frac{13}{48} = 13\). 2. For b), distribute: \(48 \cdot \frac{7}{12} - 48 \cdot \frac{5}{16} = 28 - 15 = 13\). 3. Both methods give \(13\).

Answer

a) \(13\) b) \(13\)

5122547

Apply the distributive property to evaluate each expression. Show your work. a) \(\frac{4}{5}(10 - 2.5)\) b) \(18\left(\frac{1}{2} + \frac{1}{3} - \frac{1}{6}\right)\) c) \(\frac{7}{11}(22 - 5.5)\)

Hints

- Multiply the outside factor by every term inside the parentheses. - Keep each addition or subtraction sign. - Simplify products by canceling common factors when possible.

Solution

1. For a), distribute: \(\frac{4}{5} \cdot 10 - \frac{4}{5} \cdot 2.5 = 8 - 2 = 6\). 2. For b), distribute: \(18 \cdot \frac{1}{2} + 18 \cdot \frac{1}{3} - 18 \cdot \frac{1}{6} = 9 + 6 - 3 = 12\). 3. For c), distribute: \(\frac{7}{11} \cdot 22 - \frac{7}{11} \cdot 5.5 = 14 - 3.5 = 10.5\).

Answer

a) \(6\) b) \(12\) c) \(10.5\)

5122557

For each expression, decide whether it is more efficient to evaluate inside the parentheses first or to apply the distributive property. Evaluate using your chosen method and briefly justify it. a) \(12(100 + 3)\) b) \(15\left(\frac{1}{3} + \frac{4}{5}\right)\) c) \(6.4(13 - 3)\)

Hints

- Mentally preview both methods before choosing. - Distributing can avoid adding fractions with unlike denominators. - Evaluating parentheses first can be best when the result is \(10\) or \(100\).

Solution

1. For a), distributing is efficient because multiplying by \(100\) is easy: \(12 \cdot 100 + 12 \cdot 3 = 1200 + 36 = 1236\). 2. For b), distributing avoids adding unlike fractions first: \(15 \cdot \frac{1}{3} + 15 \cdot \frac{4}{5} = 5 + 12 = 17\). 3. For c), evaluate the parentheses first because \(13 - 3 = 10\): \(6.4 \cdot 10 = 64\).

Answer

a) \(1236\); distributing is efficient. b) \(17\); distributing is efficient. c) \(64\); evaluating the parentheses first is efficient.

5122587

Let \(T = (-8)(12.5 - 5)\). a) Evaluate \(T\) by subtracting inside the parentheses first. b) Evaluate \(T\) by applying the distributive property. c) Which method do you find easier for mental calculation? Briefly justify your choice.

Hints

- For a), subtract inside the parentheses before multiplying. - For b), multiply \(-8\) by both terms and keep track of the signs. - Compare the intermediate calculations, not only the number of steps.

Solution

1. For a), \(12.5 - 5 = 7.5\), so \((-8)(7.5) = -60\). 2. For b), distribute: \((-8)(12.5) - (-8)(5) = -100 - (-40) = -60\). 3. Both methods are valid. One reasonable choice is the distributive method because \((-8)(12.5) = -100\) creates an easy intermediate value.

Answer

a) \(-60\) b) \(-60\) c) Answers may vary. For example, distributing may be easier because it creates the product \(-100\).

5122647

Determine whether each equation is true or false. For each false equation, correct the right side so that it is equivalent to the original expression on the left. a) \((-0.2) \cdot 15 + (-0.2) \cdot 5 = (-0.2)(15 + 5)\) b) \(\frac{3}{4} - \left(\frac{1}{2} - \frac{1}{4}\right) = \frac{3}{4} - \frac{1}{2} - \frac{1}{4}\) c) \((-3) \cdot 7 - (-3) \cdot 4 = (-3)(7 - 4)\)

Hints

- Check whether the distributive property has been applied correctly. - A subtraction sign before parentheses changes the sign of every term inside. - When uncertain, evaluate both sides separately.

Solution

1. For a), factoring out \(-0.2\) gives \((-0.2)(15 + 5)\), so the equation is true. 2. For b), subtracting the grouped difference changes both signs: \(\frac{3}{4} - \frac{1}{2} + \frac{1}{4}\). The given right side is false. 3. For c), factoring out \(-3\) gives \((-3)(7 - 4)\), so the equation is true.

Answer

a) True. b) False. Corrected right side: \(\frac{3}{4} - \frac{1}{2} + \frac{1}{4}\). c) True.

5122695

Write an expression for each description, then evaluate it efficiently. a) Subtract the difference of \(7.8\) and \(2.4\) from the difference of \(10.2\) and \(2.4\). b) Multiply the difference of \(25\) and \(2.5\) by \(4\).

Hints

- Translate “subtract ... from ...” carefully so the order is correct. - Use parentheses to preserve each described difference. - After writing the expressions, look for cancellation or distribution that simplifies the arithmetic.

Solution

1. For a), write \((10.2-2.4)-(7.8-2.4)\). Removing the second parentheses gives \(10.2-2.4-7.8+2.4=10.2-7.8=2.4\). 2. For b), write \((25-2.5)\times4\). Distributing gives \(25\times4-2.5\times4=100-10=90\).

Answer

a) Expression: \((10.2-2.4)-(7.8-2.4)\); value: \(2.4\) b) Expression: \((25-2.5)\times4\); value: \(90\)

5122727

A shopper uses a \(\$100\) gift card for an online order. The cart contains a game that costs \(\$24.99\), headphones that cost \(\$35.50\), and shipping that costs \(\$x\). a) Write and simplify an expression for the amount left on the gift card. b) Find the amount left when shipping costs \(\$5.95\).

Hints

- Combine the fixed costs first. - The variable \(x\) represents the shipping charge. - A larger shipping charge leaves a smaller balance.

Solution

1. Subtract all costs from the gift-card balance: \(100-(24.99+35.50+x)\). 2. Combine the fixed costs: \(24.99+35.50=60.49\). Therefore, \(100-(60.49+x)=39.51-x\). 3. For \(x=5.95\), \(39.51-5.95=33.56\).

Answer

a) \(100-(24.99+35.50+x)\), or \(39.51-x\) b) \(\$33.56\)

5122737

A day has \(24\) hours. A student plans \(8.5\) hours for sleep, \(6.5\) hours for school, \(1.5\) hours for homework, and \(t_H\) hours for hobbies. Let \(F\) represent the remaining free time. a) Write and simplify an expression for \(F\) in terms of \(t_H\). b) School time increases to \(7.5\) hours, while sleep decreases to \(8.0\) hours. How does this change \(F\)? Justify your answer by comparing the expressions.

Hints

- Add all fixed time amounts first. - Increasing the total time used for other activities decreases the free time. - Write a new expression with the changed values, then compare it with the original expression.

Solution

1. The original free time is \(F=24-(8.5+6.5+1.5+t_H)\). 2. Combine the fixed times: \(8.5+6.5+1.5=16.5\). Thus, \(F=24-16.5-t_H=7.5-t_H\). 3. With the changes, \(F_{\text{new}}=24-(8.0+7.5+1.5+t_H)\). 4. Since \(8.0+7.5+1.5=17.0\), \(F_{\text{new}}=7.0-t_H\). This is \(0.5\) hour less than the original expression.

Answer

a) \(F=7.5-t_H\) b) The free time decreases by \(0.5\) hour because the fixed-time total increases from \(16.5\) hours to \(17.0\) hours.

5122967

Two students evaluate \(-8\left(\frac{1}{4} - \frac{3}{8}\right)\). Lucas subtracts inside the parentheses first. Sarah applies the distributive property first. Show both methods, then decide which is more efficient for this expression. Justify your choice.

Hints

- Carry out the subtraction inside the parentheses for one method. - For the other method, multiply \(-8\) by both terms and keep the subtraction sign. - Compare the kinds of intermediate values each method creates.

Solution

1. Lucas’s method: \(\frac{1}{4} - \frac{3}{8} = \frac{2}{8} - \frac{3}{8} = -\frac{1}{8}\). 2. Then \(-8\left(-\frac{1}{8}\right) = 1\). 3. Sarah’s method: \(-8 \cdot \frac{1}{4} - \left(-8 \cdot \frac{3}{8}\right) = -2 - (-3) = 1\). 4. Both methods are efficient. Distributing may be more convenient because \(-8\) cancels with both denominators and produces whole-number products.

Answer

Both methods give \(1\). Distributing is especially efficient because the factor \(-8\) cancels with both denominators.

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 \(\frac{8}{3}\)

5123067

Simplify each expression first. Then evaluate it for the given variable value. a) \(T(x)=15x-4.5x+0.5x\) for \(x=1.2\) b) \(T(y)=y(12-5.5)-1.5y\) for \(y=\frac{4}{5}\)

Hints

- Simplify each expression before substituting the variable value. - Combine coefficients of like terms. - In part b, simplify the expression inside parentheses first.

Solution

1. For part a, combine the coefficients of \(x\): \(15x-4.5x+0.5x=(15-4.5+0.5)x=11x\). 2. Substitute \(x=1.2\): \(11\cdot1.2=13.2\). 3. For part b, simplify inside the parentheses and combine like terms: \(y(12-5.5)-1.5y=6.5y-1.5y=5y\). 4. Substitute \(y=\frac{4}{5}\): \(5\cdot\frac{4}{5}=4\).

Answer

a) \(13.2\) b) \(4\)

5124377

Consider the expressions \(A = 5x - 15\) and \(B = 5(x - 3)\). a) Evaluate both expressions for \(x = 10\), \(x = 0\), and \(x = -2\). b) What do you notice about the results? Justify your observation using a property of operations.

Hints

- Evaluate the expression inside the parentheses before multiplying. - Compare the two results for each input. - Which property lets you rewrite a product involving parentheses?

Solution

1. For \(x = 10\), \(A = 5 \cdot 10 - 15 = 35\) and \(B = 5(10 - 3) = 35\). 2. For \(x = 0\), \(A = 5 \cdot 0 - 15 = -15\) and \(B = 5(0 - 3) = -15\). 3. For \(x = -2\), \(A = 5 \cdot (-2) - 15 = -25\) and \(B = 5(-2 - 3) = -25\). 4. By the distributive property, \(5(x - 3) = 5x - 15\). Therefore, the expressions are equivalent and have the same value for every \(x\).

Answer

a) At \(x = 10\), both values are \(35\); at \(x = 0\), both values are \(-15\); at \(x = -2\), both values are \(-25\). b) The expressions always have the same value because the distributive property gives \(5(x - 3) = 5x - 15\).

5124627

Two expressions are equivalent if they have the same value for every value of the variable. Simplify the expressions to determine whether \(T_1\) and \(T_2\) are equivalent. \(T_1 = 4(2z - 3) - (5z - 8)\) \(T_2 = 3(z - 2) + 2\)

Hints

- Simplify each expression separately. - If both simplify to exactly the same expression, they are equivalent. - In \(T_1\), be especially careful when subtracting \((5z - 8)\).

Solution

1. Simplify \(T_1\): \(4(2z - 3) - (5z - 8) = 8z - 12 - 5z + 8 = 3z - 4\). 2. Simplify \(T_2\): \(3(z - 2) + 2 = 3z - 6 + 2 = 3z - 4\). 3. Both expressions simplify to \(3z - 4\), so they are equivalent.

Answer

Yes. The expressions are equivalent because both simplify to \(3z - 4\).

5124637

Complete the table, then simplify the expressions to determine which are equivalent. Expression \(A=3(x+4)\) Expression \(B=3x+4\) Expression \(C=5x+12-2x\) <table> <tr> <td>\(x\)</td> <td>\(0\)</td> <td>\(2\)</td> <td>\(5\)</td> </tr> <tr> <td>Expression \(A\)</td> <td></td> <td></td> <td></td> </tr> <tr> <td>Expression \(B\)</td> <td></td> <td></td> <td></td> </tr> <tr> <td>Expression \(C\)</td> <td></td> <td></td> <td></td> </tr> </table> Which expressions are equivalent? Justify your answer by simplifying them.

Hints

- Substitute each given value of \(x\) carefully. - Use the distributive property to remove parentheses. - Combine like terms and compare the simplified expressions.

Solution

1. For \(x=0\): \(A=3\cdot(0+4)=12\), \(B=3\cdot0+4=4\), and \(C=5\cdot0+12-2\cdot0=12\). 2. For \(x=2\): \(A=3\cdot(2+4)=18\), \(B=3\cdot2+4=10\), and \(C=5\cdot2+12-2\cdot2=18\). 3. For \(x=5\): \(A=3\cdot(5+4)=27\), \(B=3\cdot5+4=19\), and \(C=5\cdot5+12-2\cdot5=27\). 4. Simplify \(A\): \(3(x+4)=3x+12\). 5. Simplify \(C\): \(5x+12-2x=3x+12\). 6. Expressions \(A\) and \(C\) are equivalent because they simplify to the same expression. Expression \(B\) is not equivalent to them.

Answer

<table> <tr><td>\(x\)</td><td>\(0\)</td><td>\(2\)</td><td>\(5\)</td></tr> <tr><td>Expression \(A\)</td><td>\(12\)</td><td>\(18\)</td><td>\(27\)</td></tr> <tr><td>Expression \(B\)</td><td>\(4\)</td><td>\(10\)</td><td>\(19\)</td></tr> <tr><td>Expression \(C\)</td><td>\(12\)</td><td>\(18\)</td><td>\(27\)</td></tr> </table> Expressions \(A\) and \(C\) are equivalent because both simplify to \(3x+12\).

5124647

Let \(T=12x-(4x+8)\). Simplify \(T\). Then determine which expressions are equivalent to \(T\). Justify each choice by rewriting the expression. \(T_1=8x+8\) \(T_2=4(2x-2)\) \(T_3=16x-8-8x\) \(T_4=8(x-1)\)

Hints

- Distribute the negative sign in the original expression carefully. - Rewrite each expression in the form \(ax+b\). - Use the distributive property and combine like terms.

Solution

1. Simplify the original expression: \(T=12x-(4x+8)=12x-4x-8=8x-8\). 2. \(T_1=8x+8\) is not equivalent because its constant term differs from \(8x-8\). 3. \(T_2=4(2x-2)=8x-8\), so it is equivalent. 4. \(T_3=16x-8-8x=8x-8\), so it is equivalent. 5. \(T_4=8(x-1)=8x-8\), so it is equivalent.

Answer

The simplified expression is \(T=8x-8\). The equivalent expressions are \(T_2\), \(T_3\), and \(T_4\).

5124657

Luke and Maya simplify \(6x-2(x-4)\). Luke writes: \(6x-2(x-4)=6x-2x-8=4x-8\) Maya writes: \(6x-2(x-4)=6x-2x+8=4x+8\) a) Substitute \(x=0\) into the original expression and into each student’s result. Which student could be correct? b) Explain the error in the incorrect work.

Hints

- Equivalent expressions must have the same value for every input. - Substitute \(x=0\) into all three expressions. - Check how \(-2\) multiplies both terms inside the parentheses.

Solution

1. Substitute \(x=0\) into the original expression: \(6\cdot0-2\cdot(0-4)=0-2\cdot(-4)=8\). 2. Luke’s expression gives \(4\cdot0-8=-8\). 3. Maya’s expression gives \(4\cdot0+8=8\). 4. Maya’s result matches the original expression for \(x=0\), so Maya could be correct. 5. Luke did not distribute \(-2\) correctly to \(-4\). Since \((-2)\cdot(-4)=8\), the constant term should be \(+8\), not \(-8\).

Answer

a) The original expression equals \(8\) when \(x=0\). Luke’s result is \(-8\), and Maya’s result is \(8\). Maya could be correct. b) Luke made a sign error when distributing \(-2\). The product \((-2)\cdot(-4)\) is \(+8\).

5124677

A student wrote: \(12b-(5b+7)+3=7b+10\) Determine whether the result is correct. If it is not, identify the error and give the correct simplified expression.

Hints

- Focus on the negative sign before the parentheses. - Simplify the expression independently before comparing results. - Check the variable terms and constants separately.

Solution

1. Distribute the negative sign: \(12b-(5b+7)+3=12b-5b-7+3\). 2. Combine the variable terms: \(12b-5b=7b\). 3. Combine the constants: \(-7+3=-4\). 4. The correct simplified expression is \(7b-4\). 5. The student failed to change the sign of \(+7\) when removing the parentheses and incorrectly combined the constants as \(7+3\).

Answer

The result is incorrect. The \(+7\) inside the subtracted parentheses must become \(-7\). The correct simplification is \(7b-4\).

5124717

Consider these two expressions: Expression A: \(30-x-10-5\) Expression B: \(30-x-(10-5)\) a) Simplify each expression. b) Find \(\text{Expression B}-\text{Expression A}\). Does the result depend on the value of \(x\)? Explain.

Hints

- Combine the constant terms in each expression. - Pay attention to the subtraction sign before the parentheses in Expression B. - When subtracting the expressions, distribute the subtraction sign to both terms in the second expression.

Solution

1. Simplify Expression A by combining the constants: \(30-x-10-5=15-x\). 2. Simplify Expression B by evaluating the parentheses: \(30-x-(10-5)=30-x-5=25-x\). 3. Subtract the expressions: \((25-x)-(15-x)=25-x-15+x=10\). 4. The \(x\)-terms cancel, so the difference is always \(10\), regardless of the value of \(x\).

Answer

a) Expression A: \(15-x\); Expression B: \(25-x\) b) The difference is \(10\). It does not depend on \(x\) because the \(x\)-terms cancel.

5124797

Sort the expressions into two groups of three equivalent expressions. 1. \(6a-12+2a\) 2. \(3(a+4)-a\) 3. \(4(2a-3)\) 4. \(2a+12\) 5. \(10a-12-2a\) 6. \(2(a+6)\)

Hints

- Simplify every expression before grouping them. - Distribute a factor to every term inside parentheses. - Combine like terms after distributing.

Solution

1. Expression 1 simplifies to \(6a-12+2a=8a-12\). 2. Expression 2 simplifies to \(3(a+4)-a=3a+12-a=2a+12\). 3. Expression 3 simplifies to \(4(2a-3)=8a-12\). 4. Expression 4 is already \(2a+12\). 5. Expression 5 simplifies to \(10a-12-2a=8a-12\). 6. Expression 6 simplifies to \(2(a+6)=2a+12\). 7. Therefore, expressions 1, 3, and 5 form one group, and expressions 2, 4, and 6 form the other.

Answer

Group 1, equivalent to \(8a-12\): expressions 1, 3, and 5 Group 2, equivalent to \(2a+12\): expressions 2, 4, and 6

5124907

A rectangular garden bed has width \(x\) meters. Its length is \(3\,\text{m}\) more than twice its width. a) Write an expression for the perimeter \(P\). b) Simplify the expression. c) Find the total fence length needed when the width is \(4.5\,\text{m}\).

Hints

- Express the length in terms of \(x\). - Use the perimeter formula for a rectangle. - Distribute before combining like terms.

Solution

1. The width is \(x\), and the length is \(2x+3\). 2. The perimeter is \(P=2x+2(2x+3)\). 3. Distribute and combine like terms: \(P=2x+4x+6=6x+6\). 4. For \(x=4.5\), \(P=6\cdot4.5+6=27+6=33\,\text{m}\).

Answer

a) \(P=2x+2(2x+3)\) b) \(P=6x+6\) c) \(33\,\text{m}\)

5124917

The expression \(P=4a+2(a+5)\) represents the perimeter of a geometric figure. a) Describe one possible figure and list its side lengths so that its perimeter matches the expression. b) Find the perimeter when \(a=3.5\,\text{cm}\). c) Simplify the expression.

Hints

- Treat each term as representing one or more side lengths. - How many side lengths are represented by the expression? - Use the distributive property before combining like terms.

Solution

1. One possible figure is an irregular hexagon with four sides of length \(a\) and two sides of length \(a+5\). 2. For \(a=3.5\,\text{cm}\), \(P=4\cdot3.5+2(3.5+5)=14+17=31\). The perimeter is \(31\,\text{cm}\). 3. Distribute and combine like terms: \(P=4a+2a+10=6a+10\).

Answer

a) Answers will vary. One example is a hexagon with side lengths \(a\), \(a\), \(a\), \(a\), \(a+5\), and \(a+5\). b) \(31\,\text{cm}\) c) \(P=6a+10\)

5124927

Figure A is an equilateral triangle with side length \(2s + 4\). Figure B is a square with side length \(1.5s + 3\). Use algebra to show that the two figures have the same perimeter for every value of \(s\) that gives positive side lengths.

Hints

- How many equal sides does an equilateral triangle have? How many sides does a square have? - Write a perimeter expression for each figure. - Use the distributive property to simplify both expressions. - What must be true of the simplified expressions for the perimeters always to match?

Solution

1. The triangle's perimeter is \(P_A = 3(2s + 4) = 6s + 12\). 2. The square's perimeter is \(P_B = 4(1.5s + 3) = 6s + 12\). 3. The perimeter expressions are identical, so \(P_A = P_B\) for every allowable value of \(s\). Both side lengths are positive when \(s > -2\).

Answer

Both perimeter expressions simplify to \(6s + 12\), so the figures have equal perimeters for every \(s > -2\).

5125017

Determine whether these two rules always produce the same result for any number \(x\). Justify your answer by writing and simplifying expressions. Rule A: Add \(4\) to a number, then multiply the entire result by \(3\). Rule B: Add twice the number to the number itself, then add \(12\).

Hints

- Write a separate algebraic expression for each rule. - In Rule A, the phrase “the entire result” means the sum should be inside parentheses. - Expand any parentheses and combine like terms. - How can simplified expressions show that two rules always agree?

Solution

1. Rule A is represented by \(T_A = 3(x + 4)\). Using the distributive property, \(T_A = 3x + 12\). 2. Rule B is represented by \(T_B = 2x + x + 12\). Combining like terms gives \(T_B = 3x + 12\). 3. Since both rules simplify to \(3x + 12\), they always produce the same result.

Answer

Yes. Rule A gives \(3(x + 4) = 3x + 12\), and Rule B gives \(2x + x + 12 = 3x + 12\). The expressions are equivalent.

5125067

For a school carnival, a seventh-grade class compares two drink suppliers. Lakeview Drinks charges \(\$15.00\) for delivery and \(\$0.80\) per bottle. Riverbend Bottling charges \(\$5.00\) for delivery and \(\$1.20\) per bottle. a) Write an expression for the total cost from each supplier for \(x\) bottles. b) Find the cost of \(30\) bottles from each supplier. c) Two classes order the same number, \(x\), of bottles. One orders from Lakeview Drinks and the other from Riverbend Bottling. A student claims their combined cost is \(20+2x\) dollars. Verify the claim by combining the expressions.

Hints

- The delivery charge is a one-time fixed cost. - When adding expressions, combine only like terms. - Check the combined expression by substituting a value for \(x\).

Solution

1. Lakeview Drinks' cost is \(15+0.80x\). Riverbend Bottling's cost is \(5+1.20x\). 2. For \(x=30\), Lakeview Drinks costs \(15+0.80\cdot30=39\), and Riverbend Bottling costs \(5+1.20\cdot30=41\). 3. Add the expressions: \((15+0.80x)+(5+1.20x)=20+2x\). The student's expression is correct.

Answer

a) Lakeview Drinks: \(15+0.80x\); Riverbend Bottling: \(5+1.20x\) b) Lakeview Drinks: \(\$39.00\); Riverbend Bottling: \(\$41.00\) c) Yes. \((15+0.80x)+(5+1.20x)=20+2x\).

5126057

Let \(T_1=4n+8\). A student claims that \(T_2=4(n+8)\) is equivalent to \(T_1\). 1. Test the claim by evaluating both expressions for \(n=2\). 2. Explain the error using the distributive property. 3. How should the expression inside the parentheses in \(T_2\) be changed so that it is equivalent to \(T_1\)?

Hints

- Substitute \(n=2\) into both expressions. - Distribute the factor outside the parentheses to every term inside. - Find the number that becomes \(8\) after multiplication by \(4\).

Solution

1. For \(n=2\), \(T_1=4\cdot2+8=16\), while \(T_2=4\cdot(2+8)=40\). Since \(16\ne40\), the claim is false. 2. By the distributive property, \(4(n+8)=4n+32\), not \(4n+8\). The factor \(4\) must multiply both terms inside the parentheses. 3. The number inside the parentheses must produce \(8\) when multiplied by \(4\). Since \(8\div4=2\), the equivalent expression is \(4(n+2)\).

Answer

1. \(T_1=16\) and \(T_2=40\), so the claim is false. 2. \(4(n+8)=4n+32\), not \(4n+8\). 3. Change \(T_2\) to \(4(n+2)\).

5126067

Two expressions are given: \(A=5x-(2x+3)\) \(B=3x-3\) 1. Simplify \(A\) step by step to show that the expressions are equivalent for every value of \(x\). 2. Evaluate both expressions for \(x=4.5\).

Hints

- Distribute the negative sign before combining like terms. - Compare the simplified form of \(A\) with \(B\). - Substitute the value of \(x\) after simplifying.

Solution

1. Distribute the negative sign: \(A=5x-2x-3\). 2. Combine like terms: \(A=3x-3\), which is identical to \(B\). Therefore, the expressions are equivalent. 3. For \(x=4.5\), \(3\cdot4.5-3=13.5-3=10.5\). Both expressions have this value.

Answer

1. \(A=5x-2x-3=3x-3=B\). 2. For \(x=4.5\), both expressions equal \(10.5\).

5126267

There are three piles of counters. The left and right piles each start with exactly \(12\) counters. The middle pile starts with \(z\) counters. Follow these steps: 1. Move \(5\) counters from the left pile to the middle pile. 2. Move \(3\) counters from the right pile to the middle pile. 3. Count the counters remaining in the left pile. Remove that same number of counters from the middle pile and set them aside. a) How many counters remain in the left pile after step \(1\)? b) Write and simplify an expression for the number of counters in the middle pile after step \(3\). c) How many counters are in the middle pile at the end if \(z=10\)?

Hints

- Track the number of counters in each pile after each step. - First find the fixed number left in the left pile. - In step \(3\), subtract that fixed number from the middle-pile expression.

Solution

1. After step \(1\), the left pile has \(12-5=7\) counters. 2. After steps \(1\) and \(2\), the middle pile has \(z+5+3=z+8\) counters. 3. Step \(3\) removes \(7\) counters from the middle pile, so the final amount is \((z+8)-7=z+1\). 4. When \(z=10\), the final amount is \(10+1=11\).

Answer

a) \(7\) counters b) \(z+1\) c) \(11\) counters

5128297

Determine whether the expressions are equivalent. Justify your conclusion by simplifying both. Expression 1: \(3(4x - 5) + 8\) Expression 2: \(2x + 5(2x - 1) - 2\)

Hints

- Expand the parentheses first. - Combine the terms containing \(x\) separately from the constants. - What does it mean for two expressions to be equivalent?

Solution

1. Simplify Expression 1: \(3(4x - 5) + 8 = 12x - 15 + 8 = 12x - 7\). 2. Simplify Expression 2: \(2x + 5(2x - 1) - 2 = 2x + 10x - 5 - 2 = 12x - 7\). 3. Since both simplify to \(12x - 7\), the expressions are equivalent.

Answer

Yes. The expressions are equivalent because both simplify to \(12x - 7\).

5142387

The expression \(4(x+5)+6x\) is simplified below. Step 1: \(4x+4\cdot5+6x\) Step 2: \(4x+6x+4\cdot5\) Step 3: \((4+6)x+4\cdot5\) Step 4: \(10x+20\) Name the mathematical property used in Steps 1, 2, and 3. Briefly explain how Step 2 helps with the remaining simplification.

Hints

- In Step 1, identify the property that distributes a factor across a sum. - Compare the order of the addends in Steps 1 and 2. - In Step 3, notice that \(x\) is factored out of two terms.

Solution

1. Step 1 uses the distributive property: \(4(x+5)=4x+4\cdot5\). 2. Step 2 uses the commutative property of addition to place the like terms \(4x\) and \(6x\) next to each other. 3. Step 3 uses the distributive property in reverse to factor out \(x\): \(4x+6x=(4+6)x\). 4. Step 2 groups the like terms together, making it easier to combine them in the next step.

Answer

Step 1: distributive property Step 2: commutative property of addition Step 3: distributive property Step 2 places the like terms next to each other so they can be combined easily.

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 \(0\).

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.

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

\(100\)

5222137

At a charity walk, Class A raises \(\$x\). Class B raises \(20\%\) more than Class A. Class C raises \(\$50\) less than Class B. 1. Write an expression for the amount raised by Class B. 2. Write an expression for the amount raised by Class C. 3. Write and simplify an expression for the total amount raised by all three classes.

Hints

- Write the percent increase as a decimal. - Express each class's amount separately before adding. - Combine like terms when simplifying the total. - Distinguish variable terms from fixed amounts.

Solution

1. A \(20\%\) increase gives \(x+0.20x=1.20x\). 2. Class C raises \(1.20x-50\). 3. Add the three amounts: \(x+1.20x+(1.20x-50)=3.40x-50\).

Answer

1. \(1.20x\), or \(x+0.20x\) 2. \(1.20x-50\) 3. \(3.40x-50\)

5222567

At a school-supply store, a mechanical pencil costs \(\$f\). An eraser costs \(\$k\) less than the mechanical pencil. a) Write an expression for the price of the eraser. b) Write and simplify an expression for the total price of one mechanical pencil and one eraser. c) Find the total price when \(f=8.50\) and \(k=5.65\). d) What would \(k=f\) mean for the price of the eraser?

Hints

- “\(k\) dollars less” means subtract \(k\) from the other price. - Add the two individual prices to find the total. - Combine terms with the same variable. - In part d), substitute the same value for both variables.

Solution

1. The eraser costs \(f-k\) dollars. 2. The total price is \(f+(f-k)=2f-k\). 3. Substitute the values: \(2\cdot 8.50-5.65=17.00-5.65=11.35\). 4. If \(k=f\), then the eraser price is \(f-f=0\), so the eraser would be free.

Answer

a) \(f-k\) dollars b) \(2f-k\) dollars c) \(\$11.35\) d) The eraser would cost \(\$0\), so it would be free.

5222727

In a school cafeteria, a staff lunch costs \(\$x\), and a student lunch costs \(\$y\). a) Write an expression for the total revenue when \(15\) staff members and \(120\) students buy lunch. b) Every lunch price increases by \(\$0.50\). Write and simplify a new expression for the revenue if the same numbers of people buy lunch. c) Find the revenue after the increase if the original staff lunch cost \(\$5.50\) and the original student lunch cost \(\$3.50\).

Hints

- Multiply each lunch price by the number of lunches sold. - Add \(0.50\) to each original price. - The increase applies to all \(135\) lunches. - Substitute the original prices only after writing the new expression.

Solution

1. The original revenue is \(15x+120y\). 2. The new prices are \(x+0.50\) and \(y+0.50\), so the new revenue is \(15(x+0.50)+120(y+0.50)\). 3. Distribute and combine constants: \(15x+7.50+120y+60=15x+120y+67.50\). 4. With original prices \(x=5.50\) and \(y=3.50\), the new prices are \(6.00\) and \(4.00\). The revenue is \(15\cdot 6+120\cdot 4=90+480=570\).

Answer

a) \(15x+120y\) b) \(15(x+0.50)+120(y+0.50)\), or \(15x+120y+67.50\) c) \(\$570.00\)

5224707

Determine whether each pair of expressions is equivalent. Justify each decision by simplifying. Pair A: \(T_1 = 2(3x + 4) - 2x\) and \(T_2 = 4x + 8\) Pair B: \(T_3 = 8a \div 2\) and \(T_4 = 4a\) Pair C: \(T_5 = 5 + 3b\) and \(T_6 = 8b\)

Hints

- Expand parentheses and combine like terms before comparing expressions. - Follow the order of operations. - To show that two expressions are not equivalent, one input that gives different values is enough.

Solution

1. Pair A: \(2(3x + 4) - 2x = 6x + 8 - 2x = 4x + 8\), so the pair is equivalent. 2. Pair B: \(8a \div 2 = 4a\), so the pair is equivalent. 3. Pair C: \(5\) and \(3b\) are not like terms, so \(5 + 3b\) cannot simplify to \(8b\). For example, when \(b = 0\), the values are \(5\) and \(0\). The pair is not equivalent.

Answer

Pair A: Equivalent; both expressions simplify to \(4x + 8\). Pair B: Equivalent; \(8a \div 2 = 4a\). Pair C: Not equivalent; for example, when \(b = 0\), the values are \(5\) and \(0\).

5225467

An electric company models a monthly bill as \( 18.50+0.16k, \) where \(k\) is the number of kilowatt-hours used. A billing analyst splits the usage charge into two equal components and writes \( 18.50+0.08k+0.08k. \) a) Show that the two expressions are equivalent for every value of \(k\). b) Name the property or like-term reasoning that justifies your rewrite. c) Explain what the coefficient \(0.16\) represents in the simplified form.

Hints

- Focus on the two terms containing \(k\). - Factor out \(k\) mentally from \(0.08k+0.08k\). - Add the numerical coefficients. - Equivalent expressions must have the same value for every allowed \(k\).

Solution

1. Combine the like terms in the analyst's expression: \( 0.08k+0.08k=(0.08+0.08)k=0.16k. \) 2. Therefore, \( 18.50+0.08k+0.08k=18.50+0.16k. \) 3. The rewrite uses the distributive property in reverse, together with addition of the numerical coefficients of like terms. 4. The coefficient \(0.16\) is the total charge in dollars for each kilowatt-hour used.

Answer

a) \(18.50+0.08k+0.08k=18.50+0.16k\) b) Combine like terms; equivalently, use the distributive property in reverse. c) \(0.16\) is the per-kilowatt-hour charge in dollars.

5225507

A square garden bed is an \(s \times s\) array of unit squares, where \(s\) is a positive whole number. A border one unit square wide is placed around the outside of the bed. Two students write different expressions for the number of border squares: Expression 1: \(4s+4\) Expression 2: \(4(s+1)\) a) Use both expressions to find the number of border squares when \(s=5\). b) Use algebra to determine whether the expressions are equivalent. c) Explain how the arrangement leads to \(4s+4\). What does each part of the expression represent?

Hints

- Substitute \(s=5\) into each expression. - Apply the distributive property to Expression 2. - Count the noncorner squares along the four sides, then count the corners.

Solution

1. Using Expression 1, \(4\cdot5+4=24\). 2. Using Expression 2, \(4(5+1)=24\). 3. Apply the distributive property: \(4(s+1)=4s+4\). Therefore, the expressions are equivalent. 4. The term \(4s\) counts \(s\) border squares along each of the four sides, excluding the corners. The additional \(4\) counts the four corner squares.

Answer

a) Both expressions give \(24\) border squares. b) Yes. \(4(s+1)=4s+4\). c) \(4s\) counts the noncorner border squares along the four sides, and \(+4\) counts the four corners.

5225547

Two motorboats, A and B, are \(d\) miles apart on a river and travel directly toward each other. The current moves at \(v_s\) miles per hour. Boat A travels with the current and has a still-water speed of \(v_A\) miles per hour. Boat B travels against the current and has a still-water speed of \(v_B\) miles per hour, where \(v_B>v_s\). a) Write an expression for each boat's speed relative to the shore. b) Write and simplify an expression for their closing speed \(v_c\). c) How does the current speed affect the time until the boats meet? Justify your answer using part b).

Hints

- First find each boat's speed relative to the shore. - Speeds add when objects move directly toward each other. - Remove parentheses and combine like terms. - Identify which variables remain after simplification.

Solution

1. Boat A's shore speed is \(v_A+v_s\). Boat B's shore speed is \(v_B-v_s\). 2. Because they move toward each other, add their shore speeds: \(v_c=(v_A+v_s)+(v_B-v_s)\). 3. Simplify: \(v_c=v_A+v_s+v_B-v_s=v_A+v_B\). 4. The current cancels from the closing-speed expression. Therefore, under this model, the meeting time is \(t=\frac{d}{v_A+v_B}\) and does not depend on \(v_s\).

Answer

a) Boat A: \(v_A+v_s\); Boat B: \(v_B-v_s\) b) \(v_c=v_A+v_B\) c) The current speed does not affect the meeting time because its effects cancel in the closing speed.

5227367

Suppose \(x - y = 7\). Determine the value of each expression. a) \(y - x\) b) \((x + 10) - (y + 10)\) c) \(2(y - x)\)

Hints

- Reversing a difference gives its opposite. - In b), remove the parentheses and combine the constants. - Use the result from a) to evaluate c).

Solution

1. Since \(y - x = -(x - y)\), part a) equals \(-7\). 2. Remove the parentheses in part b): \(x + 10 - y - 10 = x - y = 7\). 3. Use part a) in part c): \(2(y - x) = 2(-7) = -14\).

Answer

a) \(-7\) b) \(7\) c) \(-14\)

5229097

Three sections of temporary fencing are installed at a construction site. The first section is \(4a+3b\) feet long. The second section is \(a-b+2\) feet longer than the first section. The third section is \(2a+4b-5\) feet long. For values of \(a\) and \(b\) that make all three lengths positive, write and simplify an expression for the total length of the fencing.

Hints

- Write an expression for the second section first. - “Longer than” indicates addition. - Combine the \(a\)-terms, \(b\)-terms, and constants separately.

Solution

1. The second section has length \((4a+3b)+(a-b+2)=5a+2b+2\). 2. Add all three section lengths: \((4a+3b)+(5a+2b+2)+(2a+4b-5)\). 3. Combine like terms: \(4a+5a+2a=11a\), \(3b+2b+4b=9b\), and \(2-5=-3\). 4. The total length is \(11a+9b-3\) feet.

Answer

The total length is \(11a+9b-3\) feet.

5229127

For values of \(a\) and \(b\) that make all original and new dimensions positive, a rectangle has length \(L=4a+3b\) and width \(W=2a+b\). A second rectangle is built with a length that is \(a-b\) greater than \(L\) and a width that is \(b\) less than \(W\). Write and simplify an expression for the perimeter of the second rectangle.

Hints

- Find the new length and width separately. - “Greater than” indicates addition, and “less than” indicates subtraction. - Use \(P=2(L+W)\) for a rectangle.

Solution

1. The new length is \(L_2=(4a+3b)+(a-b)=5a+2b\). 2. The new width is \(W_2=(2a+b)-b=2a\). 3. Use the perimeter formula: \(P=2(L_2+W_2)\). 4. Substitute and simplify: \(P=2[(5a+2b)+2a]=2(7a+2b)=14a+4b\).

Answer

\(P=14a+4b\)

5229167

For values of \(a\) and \(b\) that make all side lengths positive and form a valid triangle, the first two sides are Side 1: \(a+4b\) Side 2: \(3a-2b+5\). The third side is \(2a+b+3\) less than the sum of the first two sides. Write and simplify an expression for the perimeter \(P\).

Hints

- Add the first two side lengths first. - “Less than” means subtract the entire given expression. - Add all three side lengths to find the perimeter.

Solution

1. Add the first two sides: \((a+4b)+(3a-2b+5)=4a+2b+5\). 2. Find the third side: \((4a+2b+5)-(2a+b+3)=2a+b+2\). 3. Add all three sides: \(P=(4a+2b+5)+(2a+b+2)=6a+3b+7\).

Answer

\(P=6a+3b+7\)

5229607

Consider \(T_1 = 2(a - b)\), \(T_2 = 2a - b\), and \(T_3 = 2a - 2b\). First evaluate all three expressions for each set of values and make a conjecture. Then prove by simplifying which expressions are equivalent. a) \(a = 5\), \(b = 3\) b) \(a = 1.5\), \(b = 2\) c) \(a = \frac{1}{2}\), \(b = \frac{1}{4}\)

Hints

- Substitute each pair of values into all three expressions. - Use the distributive property on \(T_1\). - In \(T_3\), both variables are multiplied by \(2\). - Compare the three results for each part.

Solution

1. For \(a = 5\) and \(b = 3\): \(T_1 = 4\), \(T_2 = 7\), and \(T_3 = 4\). 2. For \(a = 1.5\) and \(b = 2\): \(T_1 = -1\), \(T_2 = 1\), and \(T_3 = -1\). 3. For \(a = \frac{1}{2}\) and \(b = \frac{1}{4}\): \(T_1 = \frac{1}{2}\), \(T_2 = \frac{3}{4}\), and \(T_3 = \frac{1}{2}\). 4. The results suggest that \(T_1\) and \(T_3\) are equivalent. The distributive property proves it: \(2(a - b) = 2a - 2b\). The values above provide counterexamples showing that \(T_2\) is not equivalent to them.

Answer

a) \(T_1 = 4\), \(T_2 = 7\), \(T_3 = 4\) b) \(T_1 = -1\), \(T_2 = 1\), \(T_3 = -1\) c) \(T_1 = \frac{1}{2}\), \(T_2 = \frac{3}{4}\), \(T_3 = \frac{1}{2}\) \(T_1\) and \(T_3\) are equivalent because \(2(a - b) = 2a - 2b\). \(T_2\) is not equivalent to them.

5229727

Simplify each expression. 1) \(\frac{2}{3}(6a-9b)-\frac{1}{4}(8a+12b)\) 2) \(\left(x-\frac{1}{2}y+\frac{1}{3}z\right)-\left(\frac{1}{2}x+y-\frac{2}{3}z\right)+\left(\frac{1}{4}x-\frac{1}{2}y-z\right)\)

Hints

- Distribute each factor to every term inside its parentheses. - Reverse the signs in a subtracted group. - Group terms by variable before combining coefficients.

Solution

1. Distribute both factors: \(\frac{2}{3}(6a-9b)=4a-6b\) and \(\frac{1}{4}(8a+12b)=2a+3b\). 2. Subtract and combine like terms: \((4a-6b)-(2a+3b)=2a-9b\). 3. For part 2, remove the parentheses: \(x-\frac{1}{2}y+\frac{1}{3}z-\frac{1}{2}x-y+\frac{2}{3}z+\frac{1}{4}x-\frac{1}{2}y-z\). 4. Combine coefficients: \(1-\frac{1}{2}+\frac{1}{4}=\frac{3}{4}\) for \(x\), \(-\frac{1}{2}-1-\frac{1}{2}=-2\) for \(y\), and \(\frac{1}{3}+\frac{2}{3}-1=0\) for \(z\). 5. The result for part 2 is \(\frac{3}{4}x-2y\).

Answer

1) \(2a-9b\) 2) \(\frac{3}{4}x-2y\)

5229817

Simplify the expression. \(4x-[3y-(2x+5y)+7x]-(x-2y)\)

Hints

- Work from the innermost grouping symbols outward. - A negative sign before parentheses or brackets reverses every sign inside. - Simplify inside the brackets before removing them. - Combine only terms with the same variable.

Solution

1. Simplify inside the brackets: \(3y-2x-5y+7x=5x-2y\). 2. Substitute this result into the full expression: \(4x-[5x-2y]-(x-2y)\). 3. Remove the brackets and parentheses: \(4x-5x+2y-x+2y\). 4. Combine like terms to obtain \(-2x+4y\).

Answer

\(-2x+4y\)

5229867

Consider the expression \(20u-15v+10w-5\). a) Rewrite it in the form \((\dots)-(\dots)\), with \(20u\) and \(10w\) in the first set of parentheses. b) Explain why the sign of \(5\) inside the second set of parentheses differs from its sign in the original expression.

Hints

- Decide how the terms not in the first group must appear after a negative sign is distributed. - Expand your rewritten expression to check whether it matches the original. - Remember that a negative sign outside parentheses changes every sign inside.

Solution

1. Group \(20u\) and \(10w\) in the first set of parentheses: \((20u+10w)\). 2. The remaining terms are \(-15v\) and \(-5\). 3. To place a subtraction sign before the second group, write \(-(15v+5)\). Distributing the negative sign reproduces \(-15v-5\). 4. Therefore, the expression is \((20u+10w)-(15v+5)\). 5. The \(5\) is positive inside the second group because the negative sign outside the parentheses changes it to \(-5\) when distributed.

Answer

a) \((20u+10w)-(15v+5)\) b) The negative sign outside the parentheses changes \(+5\) to \(-5\) when the parentheses are removed, preserving the original expression.

5229927

Let \(u=3r-5s+2t\), \(v=-r+4s-3t\), and \(w=2r-s+t\). Simplify each expression. 1) \(u-v+w\) 2) \(-(u+v)-w\)

Hints

- Consider simplifying the expression inside parentheses before applying the negative sign. - Opposite terms, such as \(t\) and \(-t\), cancel. - Rewrite the full expression after each step to keep track of signs.

Solution

1. Substitute the expressions: \((3r-5s+2t)-(-r+4s-3t)+(2r-s+t)\). Remove the parentheses and combine like terms: \(3r-5s+2t+r-4s+3t+2r-s+t=6r-10s+6t\). 2. First find \(u+v=(3r-5s+2t)+(-r+4s-3t)=2r-s-t\). Then \(-(u+v)-w=-(2r-s-t)-(2r-s+t)\). Remove the parentheses and combine like terms: \(-2r+s+t-2r+s-t=-4r+2s\).

Answer

1) \(6r-10s+6t\) 2) \(-4r+2s\)

5230027

Consider \(T=5x+3y-[3x-(y-2x)]\). a) Simplify \(T\). b) Use the simplified expression to explain why the value of \(T\) does not depend on \(x\). c) Evaluate \(T\) for \(x=100\) and \(y=-2.5\).

Hints

- After removing the grouping symbols, check what happens to the \(x\)-terms. - Look at which variables remain in the simplified expression. - Decide whether you need to substitute \(x=100\) after simplifying.

Solution

1. Remove the inner parentheses: \(T=5x+3y-[3x-y+2x]\). 2. Combine like terms inside the brackets: \(T=5x+3y-[5x-y]\). 3. Remove the brackets and combine like terms: \(T=5x+3y-5x+y=4y\). 4. The \(x\)-terms cancel, so changing \(x\) does not affect the value of \(T\). 5. For \(y=-2.5\), \(T=4\cdot(-2.5)=-10\).

Answer

a) \(4y\) b) The \(x\)-terms cancel, so the simplified expression contains no \(x\). c) \(-10\)

5230047

A rectangle has positive side lengths \(a\) and \(b\), with \(0<x<b\). 1. Write an expression for the original perimeter \(P_{\text{old}}\). 2. The side of length \(a\) is increased by \(x\), while the side of length \(b\) is decreased by \(x\). Write and fully simplify the new perimeter \(P_{\text{new}}\). 3. Compare the two perimeter expressions. 4. Check both perimeters for \(a=40\,\text{m}\), \(b=25\,\text{m}\), and \(x=5\,\text{m}\).

Hints

- Start with the perimeter formula for a rectangle. - Replace the original side lengths with \(a+x\) and \(b-x\). - Simplify the expression inside the parentheses first. - Observe what happens to the \(x\)-terms.

Solution

1. The original perimeter is \(P_{\text{old}}=2(a+b)\). 2. The new side lengths are \(a+x\) and \(b-x\). Thus, \(P_{\text{new}}=2[(a+x)+(b-x)]\). 3. Inside the brackets, \(a+x+b-x=a+b\), so \(P_{\text{new}}=2(a+b)=P_{\text{old}}\). The perimeter is unchanged. 4. \(P_{\text{old}}=2\cdot(40+25)=130\,\text{m}\). The new sides are \(45\,\text{m}\) and \(20\,\text{m}\), so \(P_{\text{new}}=2\cdot(45+20)=130\,\text{m}\).

Answer

1. \(P_{\text{old}}=2(a+b)\) 2. \(P_{\text{new}}=2(a+b)\) 3. The perimeters are equal. 4. Both perimeters are \(130\,\text{m}\).

5230077

For \(a>0\), \(b>0\), and \(2b<3a\), a quadrilateral has perimeter \(P=10a+6b\). Three side lengths are described below. <ul><li>\(s_1=2a+b\)</li><li>\(s_2\) is \(a\) longer than \(s_1\).</li><li>\(s_3\) is \(2b\) shorter than \(s_2\).</li></ul> Write and simplify an expression for the fourth side length \(s_4\).

Hints

- Write the first three side lengths one at a time. - The perimeter is the sum of all four sides. - Distribute the subtraction sign carefully when removing the sum of the known sides.

Solution

1. The second side is \(s_2=(2a+b)+a=3a+b\). 2. The third side is \(s_3=(3a+b)-2b=3a-b\). 3. The sum of the first three sides is \((2a+b)+(3a+b)+(3a-b)=8a+b\). 4. Subtract from the perimeter: \(s_4=(10a+6b)-(8a+b)=2a+5b\).

Answer

\(s_4=2a+5b\)

5230087

A wire with total length \((16y+20)\,\text{cm}\) is bent into a quadrilateral. The value of \(y\) is chosen so that all side lengths are positive and form a quadrilateral. <ul><li>The first side is \((5y+4)\,\text{cm}\).</li><li>The second side is half as long as the first side.</li><li>The third side is \((3y+6)\,\text{cm}\).</li></ul> a) Write and simplify an expression for the fourth side \(s_4\). b) When \(y=10\), is the fourth side longer than the first side? Show the calculation.

Hints

- Multiply the first-side expression by \(\frac{1}{2}\) to find the second side. - Add the three known side expressions before subtracting from the total length. - Evaluate both side expressions at \(y=10\) before comparing.

Solution

1. The second side is \(\frac{1}{2}(5y+4)=2.5y+2\). 2. The sum of the three known sides is \((5y+4)+(2.5y+2)+(3y+6)=10.5y+12\). 3. The fourth side is \(s_4=(16y+20)-(10.5y+12)=5.5y+8\). 4. For \(y=10\), the first side is \(5\cdot 10+4=54\,\text{cm}\), and the fourth side is \(5.5\cdot 10+8=63\,\text{cm}\). Therefore, the fourth side is longer.

Answer

a) \(s_4=5.5y+8\) centimeters b) Yes. When \(y=10\), \(s_4=63\,\text{cm}\) and \(s_1=54\,\text{cm}\).

5230687

Consider \(T_1 = 3(2a + b) - 4(a - b)\) and \(T_2 = 2a + 7b\). Determine whether the expressions are equivalent. Justify your answer by simplifying \(T_1\).

Hints

- Equivalent expressions have the same value for every input. - Simplify the more complicated expression first. - Use the distributive property on both products. - Pay close attention to the negative factor \(-4\).

Solution

1. Expand \(T_1\): \(3(2a + b) - 4(a - b) = 6a + 3b - 4a + 4b\). 2. Combine like terms: \(6a - 4a + 3b + 4b = 2a + 7b\). 3. This is exactly \(T_2\), so the expressions are equivalent.

Answer

Yes. The expressions are equivalent because \(T_1\) simplifies to \(2a + 7b\).

5231317

Let \(x = 2a + 3\) and \(y = a - 5\). Substitute these expressions and simplify. 1) \(5x + 2y\) 2) \(3x - 4y\) 3) \(2(x - y)\)

Hints

- Use parentheses when substituting an expression for a variable. - A minus sign before parentheses changes every sign inside. - Combine the \(a\)-terms and constants separately.

Solution

1. \(5(2a + 3) + 2(a - 5) = 10a + 15 + 2a - 10 = 12a + 5\). 2. \(3(2a + 3) - 4(a - 5) = 6a + 9 - 4a + 20 = 2a + 29\). 3. \(2\left((2a + 3) - (a - 5)\right) = 2(a + 8) = 2a + 16\).

Answer

1) \(12a + 5\) 2) \(2a + 29\) 3) \(2a + 16\)

5238017

A savings balance \(K\) earns an annual interest rate of \(p\%\) for one year. Write an expression for the balance after one year. Then evaluate the expression for: 1) \(K = \$2500\), \(p = 2\) 2) \(K = \$12{,}000\), \(p = 1.4\)

Hints

- Find the interest as a percent of the original balance. - The ending balance must be greater than the beginning balance. - A growth factor can combine the original balance and the increase. - Substitute each set of values into your expression.

Solution

1. The balance after one year is \(K\left(1 + \frac{p}{100}\right)\). An equivalent expression is \(K + \frac{Kp}{100}\). 2. For case 1, \(\$2500\left(1 + \frac{2}{100}\right) = \$2500 \cdot 1.02 = \$2550\). 3. For case 2, \(\$12{,}000\left(1 + \frac{1.4}{100}\right) = \$12{,}000 \cdot 1.014 = \$12{,}168\).

Answer

Expression: \(K\left(1 + \frac{p}{100}\right)\) 1) \(\$2550\) 2) \(\$12{,}168\)

5238027

A store reduces an item’s original price \(x\) by \(r\%\). Write an expression for the sale price. Then evaluate it for: 1) \(x = \$45\), \(r = 20\) 2) \(x = \$129\), \(r = 15\)

Hints

- A discount subtracts part of the original price. - After a discount of \(r\%\), what percent of the original price remains? - Follow the order of operations when substituting values.

Solution

1. A discount of \(r\%\) leaves \(1 - \frac{r}{100}\) of the original price, so the sale price is \(x\left(1 - \frac{r}{100}\right)\). An equivalent expression is \(x - \frac{xr}{100}\). 2. For case 1, \(\$45\left(1 - \frac{20}{100}\right) = \$45 \cdot 0.80 = \$36\). 3. For case 2, \(\$129\left(1 - \frac{15}{100}\right) = \$129 \cdot 0.85 = \$109.65\).

Answer

Expression: \(x\left(1 - \frac{r}{100}\right)\) 1) \(\$36\) 2) \(\$109.65\)

5241447

Lukas and Sophie use two different calculation rules. Lukas says, “Choose a number, add \(4\), and multiply the result by \(5\).” Sophie says, “Choose the same number, double it, add \(8\), and multiply the new result by \(2.5\).” Write and simplify an expression in \(n\) for each rule. Determine whether the two rules give the same result for every starting number.

Hints

- What must be true of two expressions if the rules always give the same result? - Use parentheses around each quantity that is multiplied. - Apply the distributive property to both expressions. - Compare the simplified expressions.

Solution

1. Lukas’s rule gives \(5(n+4)=5n+20\). 2. Sophie’s rule gives \(2.5(2n+8)=5n+20\). 3. Both rules simplify to the same expression, \(5n+20\). Therefore, they give the same result for every starting number.

Answer

Lukas: \(5(n+4)=5n+20\) Sophie: \(2.5(2n+8)=5n+20\) The rules are equivalent, so they always give the same result.

5279497

A motorboat travels on a river. Its speed in still water is \(v\), and the river current is \(c\). 1) Write an expression for the difference between the boat's downstream speed and upstream speed. 2) Simplify the expression. What does the result show? 3) Check the result when \(v=22\) miles per hour and \(c=4\) miles per hour.

Hints

- Express the downstream and upstream speeds first. - Subtract the entire upstream expression. - Distribute the negative sign carefully. - Check whether one variable cancels.

Solution

1. The downstream speed is \(v+c\), and the upstream speed is \(v-c\). Their difference is \((v+c)-(v-c)\). 2. Simplifying gives \(v+c-v+c=2c\). The difference is twice the current speed and does not depend on the boat's still-water speed. 3. The downstream speed is \(22+4=26\) miles per hour, and the upstream speed is \(22-4=18\) miles per hour. Their difference is \(26-18=8\) miles per hour, which equals \(2\cdot 4=8\) miles per hour.

Answer

1) \((v+c)-(v-c)\) 2) \(2c\); the difference is twice the current speed. 3) \(8\) miles per hour

5279507

An airplane flies at a constant speed \(v\) in still air. A wind with speed \(w\) changes the airplane's ground speed. a) Write expressions for the ground speed with a tailwind, \(v_T\), and with a headwind, \(v_H\). b) Add \(v_T\) and \(v_H\), then divide by \(2\). Simplify and interpret the result. c) An airplane's ground speed is \(520\) miles per hour with a tailwind and \(440\) miles per hour with a headwind. Find the airplane's still-air speed and the wind speed.

Hints

- Add the wind speed for a tailwind and subtract it for a headwind. - Look for terms that cancel when the two expressions are added. - Simplify the sum before dividing by \(2\). - Compare either ground speed with the still-air speed to find \(w\).

Solution

1. With a tailwind, \(v_T=v+w\). With a headwind, \(v_H=v-w\). 2. Their mean is \(\frac{(v+w)+(v-w)}{2}=\frac{2v}{2}=v\). It equals the airplane's still-air speed. 3. The still-air speed is \(\frac{520+440}{2}=480\) miles per hour. 4. The wind speed is \(520-480=40\) miles per hour.

Answer

a) \(v_T=v+w\); \(v_H=v-w\) b) \(v\), the airplane's still-air speed c) Still-air speed: \(480\) miles per hour; wind speed: \(40\) miles per hour

5279577

Simplify the expression. \(15a-\{4b+[2a-(5b-3a)]-2b\}\)

Hints

- Begin with the innermost grouping symbols. - Reverse every sign when subtracting a grouped expression. - Combine only terms with the same variable. - Rewrite the entire expression after each step.

Solution

1. Simplify the innermost grouping: \(2a-(5b-3a)=2a-5b+3a=5a-5b\). 2. Simplify inside the braces: \(4b+(5a-5b)-2b=5a-3b\). 3. Subtract the expression in braces: \(15a-(5a-3b)=15a-5a+3b\). 4. Combine like terms to obtain \(10a+3b\).

Answer

\(10a+3b\)

5279609

Rewrite each expression without changing its value, and justify the sign or order change with a property of real numbers. a) Rewrite \(20-(x-10)\) so that a plus sign appears before the parentheses. b) Explain why \(a+(b-c)=a-(c-b)\) by removing parentheses and naming the property that allows the terms to be reordered. c) Rewrite \(m^2+(1-n)\) so that a minus sign appears before the parentheses, then expand both forms to verify equivalence.

Hints

- Replace subtraction by addition of the additive inverse. - Treat subtraction as addition of a negative term when naming properties. - In part b), compare the terms after parentheses are removed. - In part c), expand both forms as a verification, not as the first guess.

Solution

1. a) \(20-(x-10)=20+(-x+10)\). Subtracting an expression is adding its additive inverse. 2. b) The left side is \(a+b-c\). The right side is \(a-c+b\). By the commutative property of addition, \(b+(-c)=(-c)+b\), so the expressions are equivalent. 3. c) Since \(1-n=-(n-1)\), \( m^2+(1-n)=m^2-(n-1). \) Expanding either form gives \(m^2-n+1\).

Answer

a) \(20+(-x+10)\) b) Both sides simplify to the same terms; the commutative property of addition justifies reordering them. c) \(m^2-(n-1)\), and both forms expand to \(m^2-n+1\).

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.

5124687

Find the expression that belongs in the box so the two sides are equivalent. \(5(2n+1)-\Box=3n+8\)

Hints

- Expand the known product first. - Treat the box as an unknown expression. - Subtract the target expression from the expanded expression. - Substitute your result back into the original statement to check it.

Solution

1. Expand the known expression: \(5(2n+1)=10n+5\). 2. Let the missing expression be \(M\): \(10n+5-M=3n+8\). 3. Solve for \(M\): \(M=(10n+5)-(3n+8)\). 4. Distribute the negative sign and combine like terms: \(M=10n+5-3n-8=7n-3\). 5. Check: \(10n+5-(7n-3)=10n+5-7n+3=3n+8\).

Answer

\(7n-3\)

5124747

Consider the expressions \(T_1 = 2(3x - 4) + 5x\), \(T_2 = 12x - (x + 8)\), and \(T_3 = 11x - 4\). a) Simplify the expressions and determine which are equivalent. b) Change only the constant term in \(T_3\) so that the expression has a value of \(0\) when \(x = 2\).

Hints

- Equivalent expressions simplify to the same expression. - A minus sign before parentheses changes the sign of every term inside. - For part b), substitute \(x = 2\) into \(11x\), then determine the constant needed to make the total \(0\).

Solution

1. Simplify \(T_1\): \(2(3x - 4) + 5x = 6x - 8 + 5x = 11x - 8\). 2. Simplify \(T_2\): \(12x - (x + 8) = 12x - x - 8 = 11x - 8\). 3. Therefore, \(T_1\) and \(T_2\) are equivalent. Since \(T_3 = 11x - 4\), it is not equivalent to them. 4. At \(x = 2\), the variable term is \(11 \cdot 2 = 22\). To make the value \(0\), the constant term must be \(-22\). The revised expression is \(11x - 22\).

Answer

a) \(T_1\) and \(T_2\) are equivalent; both simplify to \(11x - 8\). \(T_3\) is not equivalent to them. b) \(11x - 22\)

5124807

A student claims that \(T_1=0.5(12y-8)+3y\) is equivalent to \(T_2=3(3y-2)-y+2\). a) Check the claim by simplifying both expressions. b) Change exactly one number in \(T_2\) so the resulting expression \(T_3\) is equivalent to \(T_1\). State \(T_3\).

Hints

- Simplify each expression separately. - Compare both the variable coefficient and the constant term. - Identify which number can be changed to correct the variable coefficient without changing the constant term.

Solution

1. Simplify \(T_1\): \(0.5(12y-8)+3y=6y-4+3y=9y-4\). 2. Simplify \(T_2\): \(3(3y-2)-y+2=9y-6-y+2=8y-4\). 3. The expressions are not equivalent because their variable coefficients differ. 4. One valid change is to replace the \(3\) multiplying \(y\) inside the parentheses with \(\frac{10}{3}\): \(T_3=3\left(\frac{10}{3}y-2\right)-y+2\). 5. Simplify to check: \(T_3=10y-6-y+2=9y-4\), so \(T_3\) is equivalent to \(T_1\).

Answer

a) The claim is false. \(T_1=9y-4\), while \(T_2=8y-4\). b) One possible answer is \(T_3=3\left(\frac{10}{3}y-2\right)-y+2\).

5128317

Find the value of \(a\) that makes the expressions equivalent: \(T_1 = 5k - 2(k - 6)\) \(T_2 = 3(k + a)\)

Hints

- Simplify the first expression completely. - Expand the second expression so the two forms can be compared. - What value of \(a\) makes the constant terms equal?

Solution

1. Simplify \(T_1\): \(5k - 2(k - 6) = 5k - 2k + 12 = 3k + 12\). 2. Expand \(T_2\): \(3(k + a) = 3k + 3a\). 3. For the expressions to be equivalent, the constant terms must match: \(3a = 12\). 4. Divide by \(3\): \(a = 4\).

Answer

\(a = 4\)

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.

5229627

What expression belongs in the parentheses? \((12m-5n+2k)-(\underline{\hspace{1cm}})=7m+n-4k\)

Hints

- Treat the missing expression as an unknown. - Subtract the target expression from the first expression. - Combine each variable separately.

Solution

1. Let the missing expression be \(T\): \((12m-5n+2k)-T=7m+n-4k\). 2. Subtract the result from the first expression: \(T=(12m-5n+2k)-(7m+n-4k)\). 3. Distribute the negative sign and combine like terms: \(T=12m-5n+2k-7m-n+4k=5m-6n+6k\).

Answer

\(5m-6n+6k\)

5229807

Simplify the expression step by step. \(10a-\{3b-[4a-(5b-2c)+(a-c)]\}\)

Hints

- Work from the innermost grouping symbols outward. - Reverse every sign when subtracting a grouped expression. - Rewrite the entire expression after removing each set of grouping symbols. - Combine only terms with the same variable part.

Solution

1. Remove the innermost parentheses: \(10a-\{3b-[4a-5b+2c+a-c]\}\). 2. Combine like terms inside the brackets: \(10a-\{3b-[5a-5b+c]\}\). 3. Subtract the bracketed expression: \(10a-\{3b-5a+5b-c\}\). 4. Combine like terms inside the braces: \(10a-\{8b-5a-c\}\). 5. Subtract the expression in braces and combine like terms: \(10a-8b+5a+c=15a-8b+c\).

Answer

\(15a-8b+c\)

5229877

Write \(4x-7y+2\) as a sum of two expressions when the first addend must be \(x-3y\).

Hints

- Represent the unknown addend with a variable or placeholder. - Subtract the known addend from the complete expression. - Distribute the negative sign before combining like terms.

Solution

1. Let the missing addend be \(M\). Then \((x-3y)+M=4x-7y+2\). 2. Subtract the given addend from the target expression: \(M=(4x-7y+2)-(x-3y)\). 3. Distribute the negative sign and combine like terms: \(M=4x-7y+2-x+3y=3x-4y+2\). 4. Therefore, \(4x-7y+2=(x-3y)+(3x-4y+2)\).

Answer

\((x-3y)+(3x-4y+2)\)

5229957

First simplify \(T=3a-[2b-(a-b)]\). Then substitute \(a=2x+y\) and \(b=x-y\), and simplify the result.

Hints

- Work from the innermost grouping symbols outward. - Simplify the expression in \(a\) and \(b\) before substituting. - After substituting, use the distributive property. - Reverse every sign when subtracting a grouped expression.

Solution

1. Simplify the inner grouping: \(3a-[2b-a+b]=3a-[3b-a]\). 2. Remove the brackets: \(3a-3b+a=4a-3b\). 3. Substitute \(a=2x+y\) and \(b=x-y\): \(4(2x+y)-3(x-y)\). 4. Distribute: \(8x+4y-3x+3y\). 5. Combine like terms: \(5x+7y\).

Answer

\(5x+7y\)

5230007

Let \(X=3a-4b\), \(Y=5a+2b\), and \(Z=-a+6b\). a) Simplify \(X-Y+Z\). b) Find an expression \(W\) such that \((X+Y)+W=Z\).

Hints

- Write the condition in part b as an equation involving \(W\). - Simplify \(X+Y\) before finding \(W\). - When subtracting an entire expression, use parentheses and reverse every sign inside.

Solution

1. For part a, substitute the expressions: \((3a-4b)-(5a+2b)+(-a+6b)\). 2. Remove the parentheses and combine like terms: \(3a-5a-a-4b-2b+6b=-3a\). 3. For part b, solve the condition \((X+Y)+W=Z\) for \(W\): \(W=Z-(X+Y)\). 4. Find \(X+Y=(3a-4b)+(5a+2b)=8a-2b\). 5. Substitute and simplify: \(W=(-a+6b)-(8a-2b)=-a-8a+6b+2b=-9a+8b\).

Answer

a) \(-3a\) b) \(W=-9a+8b\)

5240477

A cleaning concentrate is \(k\%\) active ingredient by volume. To prepare a cleaning solution, \(1\,\text{L}\) of the concentrate is mixed with \(n\,\text{L}\) of water. Assume the volumes add. a) Write an expression \(c(n)\) for the percent of active ingredient in the finished mixture. b) Rewrite your expression in the form \(k\cdot r(n)\%\). Explain what the factor \(r(n)\) reveals about how much of the original concentration remains. c) If the amount of water is doubled from \(n\) liters to \(2n\) liters, write the new concentration and use the two denominator structures to explain why doubling the water does not generally cut the concentration in half. Confirm with \(k=20\) and \(n=1\).

Hints

- Separate the amount of active ingredient from the total mixture volume. - After finding the concentration, factor out \(k\) so the remaining multiplicative factor can be interpreted. - When water doubles, identify exactly which term in the total-volume denominator changes. - Compare the structure of \(1+2n\) with twice \(1+n\) before substituting numbers.

Solution

1. One liter of concentrate contains \(\frac{k}{100}\,\text{L}\) of active ingredient, and the total volume is \((1+n)\,\text{L}\). 2. Therefore, \(c(n)=\frac{k}{1+n}\%\). 3. Rewrite as \(c(n)=k\cdot\frac{1}{n+1}\%\). The factor \(\frac{1}{n+1}\) is the fraction of the original concentration that remains after dilution. 4. Doubling the water gives \(c(2n)=\frac{k}{1+2n}\%\). The denominator changes from \(1+n\) to \(1+2n\), not to \(2(1+n)\), because the original \(1\,\text{L}\) of concentrate is not doubled. 5. With \(k=20\) and \(n=1\), \(c(1)=\frac{20}{2}\%=10\%\), while \(c(2)=\frac{20}{3}\%\approx6.67\%\), not \(5\%\).

Answer

a) \(c(n)=\frac{k}{1+n}\%\) b) \(c(n)=k\cdot\frac{1}{n+1}\%\); the factor \(\frac{1}{n+1}\) shows the fraction of the original concentration that remains. c) \(c(2n)=\frac{k}{1+2n}\%\). Doubling water does not double the full denominator \(1+n\). For \(k=20\), \(n=1\): \(10\%\) becomes approximately \(6.67\%\), not \(5\%\).

5279587

Simplify the expression. \(-(6u-2v)+\{3u-[5v-(u+3v)]-(2u-v)\}\)

Hints

- Work from the innermost grouping symbols outward. - Remember the negative sign before the first parentheses. - Group terms with the same variable before combining them. - Check every sign after removing the grouping symbols.

Solution

1. Simplify the innermost grouping: \(5v-(u+3v)=5v-u-3v=-u+2v\). 2. Continue inside the braces: \(3u-(-u+2v)=4u-2v\). 3. Subtract the remaining expression: \((4u-2v)-(2u-v)=4u-2v-2u+v=2u-v\). 4. Simplify the first term: \(-(6u-2v)=-6u+2v\). 5. Combine the two results: \((-6u+2v)+(2u-v)=-4u+v\).

Answer

\(-4u+v\)

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