Free math worksheets

- Both fractions already use the same-sized parts. - Combine the counts of tenths without changing the part size.

1. Both fractions use tenths, so the parts are the same size. 2. Add the numerators and keep the denominator: \(\frac{2+5}{10}=\frac{7}{10}\).

\(\frac{7}{10}\)

- Count the tenths represented by each color separately. - The two colors use equal-sized parts, so combine their counts.

1. The blue part covers \(\frac{2}{10}\), and the orange part covers \(\frac{3}{10}\). 2. Together they cover \(\frac{2}{10}+\frac{3}{10}=\frac{5}{10}\).

a) \(\frac{2}{10}\) and \(\frac{3}{10}\) b) \(\frac{5}{10}\)

- In each expression, the fractions already use equal-sized parts. - Combine the numerators while keeping the common denominator. - After evaluating, check whether the resulting numerator and denominator share a common factor.

1. For a), the denominator stays \(12\): \(\frac{7+2+1}{12}=\frac{10}{12}=\frac{5}{6}\). 2. For b), the denominator stays \(10\): \(\frac{9-3+2}{10}=\frac{8}{10}=\frac{4}{5}\). 3. For c), the denominator stays \(8\): \(\frac{7+3-4}{8}=\frac{6}{8}=\frac{3}{4}\).

a) \(\frac{5}{6}\) b) \(\frac{4}{5}\) c) \(\frac{3}{4}\)

- Rename the target amount using eighths. - Compare the target number of eighth-sized parts with the amount already used. - Each additional note contributes one eighth-sized part.

1. Rename the target in eighths: \(\frac{3}{4}=\frac{6}{8}\). 2. The dotted quarter note uses \(\frac{3}{8}\), so the remaining amount is \(\frac{6}{8}-\frac{3}{8}=\frac{3}{8}\). 3. Each eighth note is \(\frac{1}{8}\), so three eighth notes fill the remaining \(\frac{3}{8}\).

\(3\) eighth notes

- A forward move adds sixths; a backward move subtracts sixths. - All fractions already have the same denominator. - Keep the final position in sixths, as requested.

1. Represent the forward move by addition and the backward move by subtraction: \(\frac{5}{6}+\frac{2}{6}-\frac{3}{6}\). 2. Add first: \(\frac{5}{6}+\frac{2}{6}=\frac{7}{6}\). 3. Then subtract: \(\frac{7}{6}-\frac{3}{6}=\frac{4}{6}\). 4. Therefore, the marker finishes at \(\frac{4}{6}\).

\(\frac{5}{6}+\frac{2}{6}-\frac{3}{6}=\frac{4}{6}\)

- Identify the two parts and the whole. - Addition combines the parts. - Subtraction starts with the whole and removes one part.

1. The parts combine to the whole: \(\frac{3}{4}+\frac{1}{4}=1\). 2. Reversing the addends gives \(\frac{1}{4}+\frac{3}{4}=1\). 3. Subtracting either part from the whole gives the other part: \(1-\frac{3}{4}=\frac{1}{4}\) and \(1-\frac{1}{4}=\frac{3}{4}\).

\(\frac{3}{4}+\frac{1}{4}=1\) \(\frac{1}{4}+\frac{3}{4}=1\) \(1-\frac{3}{4}=\frac{1}{4}\) \(1-\frac{1}{4}=\frac{3}{4}\)

- Evaluate each expression before comparing them. - Keep twelfths as the common unit. - For part b), subtract the smaller result from the larger result.

1. The subtraction expression equals \(\frac{11}{12}-\frac{5}{12}=\frac{6}{12}\). 2. The addition expression equals \(\frac{2}{12}+\frac{3}{12}=\frac{5}{12}\). 3. Since \(\frac{6}{12}>\frac{5}{12}\), the subtraction expression is greater. 4. The difference is \(\frac{6}{12}-\frac{5}{12}=\frac{1}{12}\).

a) \(\frac{11}{12}-\frac{5}{12}\) b) \(\frac{1}{12}\)

- Keep the denominator fixed while adding eighths. - Follow each requested grouping separately. - Compare the two totals before answering part c).

1. For part a), \(\left(\frac{1}{8}+\frac{3}{8}\right)+\frac{2}{8}=\frac{4}{8}+\frac{2}{8}=\frac{6}{8}\). 2. For part b), \(\frac{1}{8}+\left(\frac{3}{8}+\frac{2}{8}\right)=\frac{1}{8}+\frac{5}{8}=\frac{6}{8}\). 3. Both groupings give \(\frac{6}{8}\), so they have the same result.

a) \(\frac{6}{8}\) b) \(\frac{6}{8}\) c) Yes.

- Find each result before comparing. - Keep all results in tenths. - Match the ordered fractions back to their letters.

1. A equals \(\frac{5}{10}\). 2. B equals \(\frac{7}{10}\). 3. C equals \(\frac{8}{10}\). 4. Since \(5<7<8\), the order from least to greatest is A, B, C.

A = \(\frac{5}{10}\), B = \(\frac{7}{10}\), C = \(\frac{8}{10}\); least to greatest: A, B, C.

- Identify the addend that is the same in both expressions. - Compare only the addends that differ. - Adding the same amount to two numbers preserves the difference between them.

1. Both expressions contain the same addend, \(\frac{3}{12}\). 2. The other addends are \(\frac{5}{12}\) and \(\frac{4}{12}\). Since \(\frac{5}{12}\) is \(\frac{1}{12}\) greater, the first complete sum is also \(\frac{1}{12}\) greater. 3. Checking gives \(\frac{8}{12}\) and \(\frac{7}{12}\), whose difference is \(\frac{1}{12}\).

\(\frac{3}{12}+\frac{5}{12}\) is greater by \(\frac{1}{12}\).

- Read each card's shaded fraction from the model. - Start by considering the largest card values. - Check that your chosen sum stays below one whole.

1. The cards show A \(=\frac{1}{8}\), B \(=\frac{2}{8}\), C \(=\frac{3}{8}\), and D \(=\frac{4}{8}\). 2. To make the greatest sum with two different cards, start with the two largest values, C and D. 3. Their sum is \(\frac{3}{8}+\frac{4}{8}=\frac{7}{8}\), which is less than \(1\). 4. Any other pair replaces one of these cards with a smaller value, so no other allowed pair can have a greater sum.

Cards C and D; \(\frac{7}{8}\)

- First determine how much must be removed from the ribbon. - Rename the remaining half yard in twelfths. - Match the required cut length to one of the visual cards.

1. The amount to cut off is \(\frac{11}{12}-\frac{1}{2}\). 2. Rename \(\frac{1}{2}\) as \(\frac{6}{12}\). The cut length must be \(\frac{5}{12}\) yard. 3. Reading the models, card E represents \(\frac{5}{12}\), so card E is the correct choice.

Card E

- All three fractions use eighth-sized parts. - Think of the equation as a missing-addend fact using the numerators. - Determine how many more eighths are needed to reach the target.

1. The missing eighths and \(3\) eighths must total \(7\) eighths. 2. The missing number of eighths is \(7-3=4\). 3. Therefore, the missing fraction is \(\frac{4}{8}\).

\(4\)

- Each small step represents \(\frac{1}{8}\). - Subtract the starting value from the ending value for each ribbon. - Use eighths to compare, subtract, and add the lengths.

1. Ribbon \(D\) runs from \(\frac{2}{8}\) to \(\frac{11}{8}\), so its length is \(\frac{11}{8}-\frac{2}{8}=\frac{9}{8}=1\frac{1}{8}\). 2. Ribbon \(E\) runs from \(\frac{5}{8}\) to \(\frac{15}{8}\), so its length is \(\frac{15}{8}-\frac{5}{8}=\frac{10}{8}=\frac{5}{4}=1\frac{1}{4}\). 3. Since \(1\frac{1}{4}=1\frac{2}{8}\), ribbon \(E\) is longer by \(\frac{1}{8}\). 4. The total length is \(1\frac{1}{8}+1\frac{1}{4}=1\frac{1}{8}+1\frac{2}{8}=2\frac{3}{8}\).

a) \(1\frac{1}{8}\) b) \(1\frac{1}{4}\) c) Ribbon \(E\) is longer by \(\frac{1}{8}\). d) \(2\frac{3}{8}\)

- Count the equal intervals between consecutive whole numbers. - Read each point by counting steps from \(0\). - Subtract the smaller value from the larger value to find the distance.

1. Each interval between whole numbers is divided into thirds, so one step is \(\frac{1}{3}\). 2. Point \(A\) is at \(\frac{2}{3}\), and point \(B\) is at \(2\frac{1}{3}=\frac{7}{3}\). 3. The distance is \(\frac{7}{3}-\frac{2}{3}=\frac{5}{3}=1\frac{2}{3}\).

a) \(A=\frac{2}{3}\), \(B=2\frac{1}{3}\) b) \(1\frac{2}{3}\)

- Identify the size of the fractional parts being combined. - When equal-sized parts are added, does the size of each part change? - Check whether the fractional pieces complete another whole.

1. The fractional parts are both fifths, so their denominators stay \(5\) when they are added. 2. \(\frac{3}{5}+\frac{2}{5}=\frac{5}{5}=1\). 3. Therefore, \(1\frac{3}{5}+\frac{2}{5}=2\). Kai incorrectly added the denominators as well as the numerators.

Kai should keep the denominator \(5\). The correct sum is \(2\).

- Rewrite one whole using eighths. - Think about pairs of positive whole-number numerators that make the numerator of one whole. - Stop when listing a new pair would only reverse one you already have.

1. One whole is \(\frac{8}{8}\), so the two numerators must add to \(8\). 2. Starting with numerator \(1\) gives \(\frac{1}{8}+\frac{7}{8}=1\). 3. Continuing without reversing a pair gives \(\frac{2}{8}+\frac{6}{8}=1\), \(\frac{3}{8}+\frac{5}{8}=1\), and \(\frac{4}{8}+\frac{4}{8}=1\). 4. Any larger first numerator would only reverse a pair already listed.

\(\frac{1}{8}+\frac{7}{8}=1\), \(\frac{2}{8}+\frac{6}{8}=1\), \(\frac{3}{8}+\frac{5}{8}=1\), \(\frac{4}{8}+\frac{4}{8}=1\)

- Compare the fractional part of the mixed number with the fraction being subtracted. - If the fractional part is too small, rename one whole using fifths. - After regrouping, subtract equal-sized fifths.

1. The fractional part \(\frac{1}{5}\) is smaller than \(\frac{3}{5}\), so one whole must be regrouped before subtracting. 2. Rename \(2\frac{1}{5}\) as \(1\frac{6}{5}\). 3. Then \(1\frac{6}{5}-\frac{3}{5}=1\frac{3}{5}\). 4. Priya subtracted the fractional parts as though \(\frac{1}{5}\) were large enough, without regrouping a whole.

Priya needed to regroup one whole. The correct difference is \(1\frac{3}{5}\).

- Express the mixed-number total entirely in tenths. - Equal addends receive the same number of tenths. - Keep denominator \(10\), as required by the problem.

1. Rewrite the total as \(\frac{12}{10}\). 2. Split the twelve tenths equally between the two addends: \(12\div2=6\). 3. Each fraction is \(\frac{6}{10}\), which is equivalent to \(\frac{3}{5}\).

\(\frac{6}{10}\)

- The second ribbon is the total length minus the first ribbon’s length. - Rewrite the half as fourths before subtracting. - Include meters with the final mixed-number length.

1. Express both lengths in fourths: \(2\frac{1}{4}=\frac{9}{4}\) and \(3\frac{1}{2}=\frac{14}{4}\). 2. Subtract the original ribbon from the total length: \(\frac{14}{4}-\frac{9}{4}=\frac{5}{4}\). 3. Convert the result to a mixed number: \(\frac{5}{4}=1\frac{1}{4}\).

\(1\frac{1}{4}\,\text{m}\)

- Evaluate the student’s proposed signs before trying a replacement. - Compare the proposed result with the target numerator. - Test the other sign order from left to right.

1. Marisol’s choice gives \(\frac{8}{12}+\frac{3}{12}-\frac{2}{12}=\frac{9}{12}\), not \(\frac{7}{12}\). 2. Using subtraction first gives \(\frac{8}{12}-\frac{3}{12}=\frac{5}{12}\). 3. Adding \(\frac{2}{12}\) next gives \(\frac{5}{12}+\frac{2}{12}=\frac{7}{12}\). 4. The correct signs are \(-\) and then \(+\).

Marisol’s choice is wrong. The correct equation is \(\frac{8}{12}-\frac{3}{12}+\frac{2}{12}=\frac{7}{12}\).

- Simplify the known addition-subtraction part first. - Rewrite one whole in twelfths. - Find the numerator needed to complete the whole.

1. Combine the known terms: \(\frac{7}{12}-\frac{2}{12}=\frac{5}{12}\). 2. One whole is \(\frac{12}{12}\), so the missing amount is \(\frac{12}{12}-\frac{5}{12}=\frac{7}{12}\). 3. Therefore, \(n=7\).

\(n=7\)

- Read how many of the eight equal parts are shaded in each model. - Ask whether combining pieces changes the size of each individual piece. - Describe what changes and what stays the same in the fraction notation.

1. The first model shows \(\frac{3}{8}\), and the second shows \(\frac{2}{8}\). 2. Together they contain five eighth-sized parts, so \(\frac{3}{8}+\frac{2}{8}=\frac{5}{8}\). 3. Adding the fractions combines counts of the same-sized eighths; it does not cut each eighth into smaller parts. 4. Therefore, when two fractions have denominator \(8\), add their numerators and keep denominator \(8\).

a) \(\frac{3}{8}+\frac{2}{8}=\frac{5}{8}\) b) The pieces are already equal-sized eighths, so adding combines their counts without changing their size. c) Add the numerators and keep denominator \(8\).

- Read the starting amount and each card amount from the eighths models. - Subtract one card at a time from the same starting fraction. - Compare how the endpoints change as the card amount gets larger.

1. Model S shows \(\frac{5}{8}\). Cards A, B, and C show \(\frac{1}{8}\), \(\frac{2}{8}\), and \(\frac{3}{8}\). 2. The possible results are \(\frac{5}{8}-\frac{1}{8}=\frac{4}{8}\), \(\frac{5}{8}-\frac{2}{8}=\frac{3}{8}\), and \(\frac{5}{8}-\frac{3}{8}=\frac{2}{8}\). 3. From least to greatest, the results are \(\frac{2}{8},\frac{3}{8},\frac{4}{8}\). 4. As the amount subtracted increases by \(\frac{1}{8}\), the result decreases by \(\frac{1}{8}\).

a) A: \(\frac{4}{8}\); B: \(\frac{3}{8}\); C: \(\frac{2}{8}\) b) \(\frac{2}{8}<\frac{3}{8}<\frac{4}{8}\) c) Each increase of \(\frac{1}{8}\) in the subtracted amount decreases the result by \(\frac{1}{8}\).

- Read both addends from the two models. - Determine how many eighths model A needs to complete its next whole. - Split model B into the amount needed for that whole and the amount left over.

1. Model A shows \(1\frac{5}{8}\), and model B shows \(\frac{6}{8}\). 2. Model A needs \(\frac{3}{8}\) to reach \(2\), so decompose \(\frac{6}{8}\) as \(\frac{3}{8}+\frac{3}{8}\). 3. Then \(1\frac{5}{8}+\frac{3}{8}=2\), with \(\frac{3}{8}\) left. 4. The final sum is \(2\frac{3}{8}\).

\(\frac{6}{8}=\frac{3}{8}+\frac{3}{8}\), so the sum is \(2\frac{3}{8}\).

- Compare how each addend changes from the first expression to the second. - Equal increases and decreases leave a total unchanged. - Use the expression that is easier to combine after recognizing the equality.

1. The first addend increases by \(\frac{1}{10}\), while the second addend decreases by \(\frac{1}{10}\). 2. These opposite changes cancel, so the two sums are equal. 3. Compute either sum: \(2\frac{8}{10}+\frac{5}{10}=2\frac{13}{10}=3\frac{3}{10}\).

The sums are equal because one addend increases by \(\frac{1}{10}\) while the other decreases by \(\frac{1}{10}\). Each sum is \(3\frac{3}{10}\).

- Treat subtraction as the distance from the smaller marked value to the larger one. - Use the labeled whole-number landmarks to organize the count. - Add the three jump lengths at the end.

1. From \(1\frac{7}{8}\) to \(2\) is \(\frac{1}{8}\). 2. From \(2\) to \(3\) is \(1\). 3. From \(3\) to \(3\frac{2}{8}\) is \(\frac{2}{8}\). 4. The total difference is \(\frac{1}{8}+1+\frac{2}{8}=1\frac{3}{8}\).

\(\frac{1}{8}+1+\frac{2}{8}=1\frac{3}{8}\), so \(3\frac{2}{8}-1\frac{7}{8}=1\frac{3}{8}\).

- Read the two quantities from models A and B. - For regrouping, trade one whole for sixth-sized parts. - For the improper-fraction method, count how many sixths are in the entire mixed number.

1. Model A shows \(2\frac{1}{6}\), and model B shows \(\frac{5}{6}\). 2. For method a), regroup \(2\frac{1}{6}\) as \(1\frac{7}{6}\). Then \(1\frac{7}{6}-\frac{5}{6}=1\frac{2}{6}=1\frac{1}{3}\). 3. For method b), rewrite \(2\frac{1}{6}=\frac{13}{6}\). Then \(\frac{13}{6}-\frac{5}{6}=\frac{8}{6}=1\frac{2}{6}=1\frac{1}{3}\). 4. Both methods give the same difference, \(1\frac{1}{3}\).

a) \(1\frac{1}{3}\) b) \(1\frac{1}{3}\)

- For parts a) and b), follow the parentheses before performing the remaining operation. - For part c), track whether the final term is added or included in the subtracted group. - Compare the two results in twelfths.

1. The first expression is \(\frac{5}{12}+\frac{2}{12}=\frac{7}{12}\). 2. The second expression is \(\frac{9}{12}-\frac{6}{12}=\frac{3}{12}=\frac{1}{4}\). 3. In the first expression, \(\frac{2}{12}\) is added after the subtraction. In the second, it is included in the amount being subtracted. 4. Therefore, the first result is greater by \(\frac{7}{12}-\frac{3}{12}=\frac{4}{12}=\frac{1}{3}\).

a) \(\frac{7}{12}\) b) \(\frac{1}{4}\) c) The first expression adds \(\frac{2}{12}\), while the second subtracts it as part of the grouped amount. The first result is greater by \(\frac{1}{3}\).

- Both expressions begin at the same fraction. - Compare moving up by an amount with moving down by that same amount. - Use direct calculation only as a final check.

1. The first expression adds \(\frac{2}{10}\), while the second subtracts \(\frac{2}{10}\) from the same starting value. 2. The results therefore differ by two copies of \(\frac{2}{10}\), which is \(\frac{4}{10}=\frac{2}{5}\). 3. Verification: \(\frac{7}{10}+\frac{2}{10}=\frac{9}{10}\) and \(\frac{7}{10}-\frac{2}{10}=\frac{5}{10}\). Their difference is \(\frac{4}{10}=\frac{2}{5}\).

The addition result is greater by \(\frac{2}{5}\). Indeed, \(\frac{9}{10}-\frac{5}{10}=\frac{4}{10}=\frac{2}{5}\).

- Decide which numerator should be subtracted to make the result as large as possible. - The remaining two numerators must fill the addition positions. - Check whether reversing the two addends creates a distinct arrangement with the same value.

1. To make the result greatest, subtract the smallest numerator, \(1\). 2. The other two numerators, \(3\) and \(5\), go in the addition positions. 3. Both addend orders give the same value: \(\frac{3}{8}+\frac{5}{8}-\frac{1}{8}=\frac{7}{8}\) and \(\frac{5}{8}+\frac{3}{8}-\frac{1}{8}=\frac{7}{8}\). 4. If \(3\) or \(5\) were subtracted instead, the result would be smaller.

\(\frac{3}{8}+\frac{5}{8}-\frac{1}{8}=\frac{7}{8}\) and \(\frac{5}{8}+\frac{3}{8}-\frac{1}{8}=\frac{7}{8}\)

- Compare how both endpoints change from the first subtraction to the second. - A difference is the distance between two values on a number line. - Use the subtraction with the whole-number minuend for the final calculation.

1. Add \(\frac{1}{8}\) to both numbers in the first subtraction: \(1\frac{7}{8}+\frac{1}{8}=2\) and \(\frac{5}{8}+\frac{1}{8}=\frac{6}{8}\). 2. Increasing both the minuend and subtrahend by the same amount preserves the distance between them. 3. Therefore, the expressions are equal. Compute \(2-\frac{6}{8}=1\frac{2}{8}=1\frac{1}{4}\).

The expressions are equal because adding \(\frac{1}{8}\) to both numbers preserves their difference. Each expression equals \(1\frac{1}{4}\).

- Translate the fraction equation into a numerator difference. - Increase both numerators together to preserve that difference. - Start with the smallest positive second numerator. - Stop when the first numerator reaches the stated maximum.

1. With equal denominators, the first numerator must be \(3\) greater than the second numerator. 2. Starting with the smallest positive second numerator gives numerator pairs \((4,1)\), \((5,2)\), \((6,3)\), \((7,4)\), and \((8,5)\). 3. Each pair gives a valid equation. 4. The next pair would begin with numerator \(9\), which exceeds the stated maximum, so the list is complete.

\(\frac{4}{8}-\frac{1}{8}=\frac{3}{8}\); \(\frac{5}{8}-\frac{2}{8}=\frac{3}{8}\); \(\frac{6}{8}-\frac{3}{8}=\frac{3}{8}\); \(\frac{7}{8}-\frac{4}{8}=\frac{3}{8}\); \(\frac{8}{8}-\frac{5}{8}=\frac{3}{8}\)

- Read the starting fraction from the blue point and the tenths scale. - Apply each move independently to the same starting point. - Compare the four resulting endpoints only after all moves have been evaluated.

1. The blue point is at \(\frac{5}{10}\). 2. Adding \(\frac{1}{10}\) gives \(\frac{6}{10}\), and adding \(\frac{3}{10}\) gives \(\frac{8}{10}\). 3. Subtracting \(\frac{2}{10}\) gives \(\frac{3}{10}\), and subtracting \(\frac{4}{10}\) gives \(\frac{1}{10}\). 4. The four endpoints are distinct, so no two moves reach the same endpoint.

\(+\frac{1}{10}\rightarrow\frac{6}{10}\); \(+\frac{3}{10}\rightarrow\frac{8}{10}\); \(-\frac{2}{10}\rightarrow\frac{3}{10}\); \(-\frac{4}{10}\rightarrow\frac{1}{10}\). No two moves reach the same endpoint.

- For part a), apply the changes in the listed order and keep a running value. - Keep the final fractional part in eighths for an easy comparison. - For part b), subtract the smaller of the start and final values from the larger, then state which value is greater.

1. After step 1: \(2\frac{3}{8}+\frac{4}{8}=2\frac{7}{8}\). 2. After step 2: \(2\frac{7}{8}-\frac{7}{8}=2\). 3. After step 3: \(2+\frac{2}{8}=2\frac{2}{8}\). 4. Compare the final value with the start: \(2\frac{3}{8}-2\frac{2}{8}=\frac{1}{8}\). The final value is \(\frac{1}{8}\) less than the starting value.

a) Step 1: \(2\frac{7}{8}\); step 2: \(2\); step 3: \(2\frac{2}{8}\) b) The final value is \(\frac{1}{8}\) less than the starting value.

- Undo the machine operations in reverse order. - Use the inverse operation for each step. - Run the found input through the original machine to check it.

1. Reverse the final addition by subtracting \(\frac{1}{8}\): \(\frac{5}{8}-\frac{1}{8}=\frac{4}{8}\). 2. Reverse the first subtraction by adding \(\frac{3}{8}\): \(\frac{4}{8}+\frac{3}{8}=\frac{7}{8}\). 3. Verification: \(\frac{7}{8}-\frac{3}{8}+\frac{1}{8}=\frac{4}{8}+\frac{1}{8}=\frac{5}{8}\).

The input is \(\frac{7}{8}\). Checking forward, \(\frac{7}{8}-\frac{3}{8}+\frac{1}{8}=\frac{5}{8}\).

- Carry out each order separately from the same starting amount. - Add the two removed fractions to compare the total removed. - Distinguish changing the order of removals from reversing the entire subtraction.

1. First order: \(1\frac{2}{8}-\frac{3}{8}=\frac{7}{8}\), then \(\frac{7}{8}-\frac{4}{8}=\frac{3}{8}\). 2. Second order: \(1\frac{2}{8}-\frac{4}{8}=\frac{6}{8}\), then \(\frac{6}{8}-\frac{3}{8}=\frac{3}{8}\). 3. Both orders remove a total of \(\frac{7}{8}\) from the same starting value, so they have the same result.

Both orders give \(\frac{3}{8}\) because each removes a total of \(\frac{7}{8}\) from the same starting value.

- Keep the denominator \(8\) and focus on the two numerators. - For the three parts, make the numerator sum less than \(8\), equal to \(8\), and greater than \(8\). - Check that each addend is still a proper fraction.

1. A sum less than \(1\) is \(\frac{2}{8}+\frac{3}{8}=\frac{5}{8}\), because \(2+3<8\). 2. A sum equal to \(1\) is \(\frac{3}{8}+\frac{5}{8}=\frac{8}{8}=1\), because \(3+5=8\). 3. A sum greater than \(1\) is \(\frac{5}{8}+\frac{6}{8}=\frac{11}{8}\), because \(5+6>8\). 4. With a common denominator of \(8\), compare the numerator sum with \(8\): less than, equal to, or greater than \(8\) determines the result’s relation to \(1\).

a) \(\frac{2}{8}+\frac{3}{8}=\frac{5}{8}\) b) \(\frac{3}{8}+\frac{5}{8}=1\) c) \(\frac{5}{8}+\frac{6}{8}=\frac{11}{8}\) Compare the numerator sum with \(8\) to determine whether the result is less than, equal to, or greater than \(1\).

- Bound each mixed number with nearby halves or whole numbers. - Use the bounds to place the difference in a one-unit interval. - Regroup only after making the estimate.

1. Since \(4\frac{1}{6}>4\) and \(2\frac{5}{6}<3\), the difference is greater than \(1\). 2. Since \(4\frac{1}{6}<4\frac{1}{2}\) and \(2\frac{5}{6}>2\frac{1}{2}\), the difference is less than \(2\). 3. Regroup: \(4\frac{1}{6}=3\frac{7}{6}\). 4. Then \(3\frac{7}{6}-2\frac{5}{6}=1\frac{2}{6}=1\frac{1}{3}\), which matches the estimate.

The difference is between \(1\) and \(2\): the whole-number bounds show it is greater than \(1\), and the half-unit bounds show it is less than \(2\). The exact difference is \(1\frac{1}{3}\).

- Combine the numerator counts while keeping eighths as the unit. - Compare the resulting numerator with \(8\). - Include every allowed value of \(n\) in exactly one category.

1. The sum is \(\frac{5+n}{8}\). 2. It is proper when \(5+n<8\), which occurs for \(n=0, 1, 2\). 3. It equals \(1\) when \(5+n=8\), so \(n=3\). 4. It is greater than \(1\) when \(5+n>8\), which occurs for \(n=4, 5, 6, 7, 8\).

Proper: \(n=0, 1, 2\) Exactly \(1\): \(n=3\) Greater than \(1\): \(n=4, 5, 6, 7, 8\)

- Read each colored part as a fraction of the twelve equal sections. - Subtract the larger, familiar benchmark amount first. - Check that the two colored parts together equal the entire shaded amount being removed.

1. The blue part is \(\frac{6}{12}=\frac{1}{2}\), and the orange part is \(\frac{1}{12}\). Together they make \(\frac{7}{12}\). 2. Subtract the blue part first: \(1-\frac{1}{2}=\frac{1}{2}=\frac{6}{12}\). 3. Then subtract the orange part: \(\frac{6}{12}-\frac{1}{12}=\frac{5}{12}\). 4. The two removed parts total \(\frac{7}{12}\), so this is the same as \(1-\frac{7}{12}\).

\(1-\frac{1}{2}=\frac{6}{12}\), then \(\frac{6}{12}-\frac{1}{12}=\frac{5}{12}\). The final result is \(\frac{5}{12}\).

- Subtract the amount used from the starting ribbon length. - Compare the two fractional parts before subtracting. - If needed, regroup one whole as fifth-sized parts.

1. Subtract the used length from the starting length: \(3\frac{2}{5}-1\frac{4}{5}\). 2. Regroup \(3\frac{2}{5}\) as \(2\frac{7}{5}\). 3. Then \(2\frac{7}{5}-1\frac{4}{5}=1\frac{3}{5}\). 4. Therefore, \(1\frac{3}{5}\) yards of ribbon remains.

\(3\frac{2}{5}-1\frac{4}{5}=1\frac{3}{5}\); \(1\frac{3}{5}\) yards remains.

- Begin with an amount that can decrease and then increase. - Describe the removal before the later addition. - Check that the situation’s final amount matches the expression.

1. One valid situation is: “A water bottle is \(\frac{10}{12}\) full, Sam drinks \(\frac{4}{12}\) of the bottle, and then adds \(\frac{1}{12}\) of the bottle.” 2. After drinking, the bottle is \(\frac{10}{12}-\frac{4}{12}=\frac{6}{12}\) full. 3. After adding water, it is \(\frac{6}{12}+\frac{1}{12}=\frac{7}{12}\) full. 4. Other situations are valid if they represent the same ordered changes.

One possible situation: A water bottle is \(\frac{10}{12}\) full, Sam drinks \(\frac{4}{12}\) of the bottle, and then adds \(\frac{1}{12}\) of the bottle. The final amount is \(\frac{7}{12}\).

- Read the starting amount from the tenths model. - Apply the two changes in the order stated. - Compare Leo's value with the amount after both changes have been made.

1. The model shows an initial amount of \(\frac{6}{10}\). 2. After Leo drinks, \(\frac{6}{10}-\frac{2}{10}=\frac{4}{10}\) remains. 3. He then adds \(\frac{1}{10}\), giving \(\frac{4}{10}+\frac{1}{10}=\frac{5}{10}\). 4. Leo stopped after the subtraction and ignored the final addition, so his claim is incorrect.

Leo is not correct. The final amount is \(\frac{5}{10}\) of the bottle.

- Name the size of one part before subtracting. - Subtraction changes how many equal parts remain, not how each part is partitioned. - Simplify only after writing the correct like-denominator difference.

1. Both fractions count parts of size \(\frac{1}{12}\), so subtraction removes three twelfth-sized parts from nine twelfth-sized parts. 2. The size of each remaining part does not change; only the number of parts changes. 3. Six twelfth-sized parts remain: \(\frac{9}{12}-\frac{3}{12}=\frac{6}{12}=\frac{1}{2}\).

The denominator remains \(12\) because the parts are still twelfth-sized. Six twelfths remain, so \(\frac{9}{12}-\frac{3}{12}=\frac{6}{12}=\frac{1}{2}\).

- Convert both the total length and the difference to twelfths. - Search for two numerators with the required total and gap. - Verify both conditions with the two ribbon lengths.

1. Rewrite the conditions in twelfths: the numerator sum is \(10\), and the numerator difference is \(2\). 2. The two whole-number numerators that add to \(10\) and differ by \(2\) are \(4\) and \(6\). 3. The ribbon lengths are \(\frac{4}{12}=\frac{1}{3}\) yard and \(\frac{6}{12}=\frac{1}{2}\) yard. 4. Their sum is \(\frac{10}{12}=\frac{5}{6}\) yard, and their difference is \(\frac{2}{12}=\frac{1}{6}\) yard.

\(\frac{4}{12}\) yard and \(\frac{6}{12}\) yard, or \(\frac{1}{3}\) yard and \(\frac{1}{2}\) yard

- The same numerator must be used in both boxes. - Think about what happens to the left side and right side as that numerator increases. - Test a value near the middle and compare the two sides.

1. Try the same numerator in both boxes while keeping all parts in tenths. 2. With \(3\) in each box, the left side is \(\frac{3}{10}+\frac{3}{10}=\frac{6}{10}\). 3. The right side is \(\frac{9}{10}-\frac{3}{10}=\frac{6}{10}\). 4. A smaller box value would make the left side smaller and the right side larger; a larger box value would do the opposite. Therefore, \(3\) is the only value that balances the equation.

\(3\); both sides equal \(\frac{6}{10}\).

- Read each card's value from the eighths model. - Find the total value of all four cards and determine what half of that total must be. - Test the possible partners for the smallest card.

1. The cards show A \(=\frac{1}{8}\), B \(=\frac{2}{8}\), C \(=\frac{3}{8}\), and D \(=\frac{4}{8}\). 2. All four cards total \(\frac{10}{8}\), so each equal pair must total \(\frac{5}{8}\). 3. A and D make \(\frac{1}{8}+\frac{4}{8}=\frac{5}{8}\). The remaining cards B and C also make \(\frac{2}{8}+\frac{3}{8}=\frac{5}{8}\). 4. If A were paired with B or C, that pair would total less than \(\frac{5}{8}\). Therefore, A must pair with D, which forces the other pair.

Pair A with D and pair B with C. Each pair totals \(\frac{5}{8}\).