- Compare the two same-size bars. - Write one half using fourths. - Then subtract the fourths.

1. The second bar shows that \(\frac{1}{2}=\frac{2}{4}\). 2. Subtract: \(\frac{3}{4}-\frac{2}{4}=\frac{1}{4}\).

\(\frac{3}{4}-\frac{1}{2}=\frac{1}{4}\).

- Rename one third using sixths. - Subtract after both fractions use the same-sized parts. - Compare the simplified difference with the benchmark.

1. Rewrite \(\frac{1}{3}=\frac{2}{6}\). 2. Subtract: \(\frac{5}{6}-\frac{2}{6}=\frac{3}{6}=\frac{1}{2}\). 3. The difference is equal to the benchmark \(\frac{1}{2}\).

\(\frac{5}{6}-\frac{1}{3}=\frac{1}{2}\), so the difference is equal to \(\frac{1}{2}\).

- How can you subtract fractions with different denominators? - What is the smallest number that is a multiple of both \(27\) and \(6\)? - When you make an equivalent fraction, multiply the numerator and denominator by the same number.

1. Find a common denominator. The least common multiple of \(27\) and \(6\) is \(54\). 2. Rewrite the fractions: \(\frac{5}{27} = \frac{10}{54}\) and \(\frac{1}{6} = \frac{9}{54}\). 3. Subtract: \(\frac{10}{54} - \frac{9}{54} = \frac{1}{54}\).

\(\frac{1}{54}\)

- What is the smallest common multiple of \(6\) and \(4\)? - How can you rewrite a fraction without changing its value? - Once the denominators match, which parts of the fractions are subtracted?

1. Find the least common denominator of \(6\) and \(4\), which is \(12\). 2. Rewrite the fractions as equivalent fractions with denominator \(12\): \(\frac{5}{6}=\frac{10}{12}\) and \(\frac{1}{4}=\frac{3}{12}\). 3. Subtract the numerators: \(\frac{10}{12}-\frac{3}{12}=\frac{7}{12}\).

Use the least common denominator \(12\), rewrite the fractions as \(\frac{10}{12}\) and \(\frac{3}{12}\), and subtract. The result is \(\frac{7}{12}\).

- Find a common denominator before subtracting fractions with different denominators. - Rewrite whole numbers as fractions when needed. - Simplify each final answer.

1. For a), use denominator \(12\): \(\frac{3}{4}-\frac{1}{6}=\frac{9}{12}-\frac{2}{12}=\frac{7}{12}\). 2. For b), use denominator \(15\): \(\frac{7}{5}-\frac{2}{3}=\frac{21}{15}-\frac{10}{15}=\frac{11}{15}\). 3. For c), rewrite \(2\) as \(\frac{16}{8}\): \(\frac{16}{8}-\frac{5}{8}=\frac{11}{8}\).

a) \(\frac{7}{12}\) b) \(\frac{11}{15}\) c) \(\frac{11}{8}\)

- Think about a denominator that can represent both fractions exactly. - Rewrite each fraction without changing its value. - After subtracting, check whether the result is already in simplest form.

1. Use denominator \(30\): \(\frac{5}{6}=\frac{25}{30}\) and \(\frac{7}{15}=\frac{14}{30}\). 2. Subtract: \(\frac{25}{30}-\frac{14}{30}=\frac{11}{30}\).

\(\frac{11}{30}\)

- Compare the total shaded length, not just the number of shaded pieces. - Ask how many fourths occupy the same length as one half. - Subtract only after both amounts are expressed in fourths.

1. Panel a) shows \(\frac{3}{4}\), and panel b) shows \(\frac{1}{2}\). 2. Panel d) has the same shaded length as panel b). Repartitioning each half into \(2\) equal pieces makes \(4\) equal parts altogether, so \(\frac{1}{2}=\frac{2}{4}\). 3. Therefore \(\frac{3}{4}-\frac{2}{4}=\frac{1}{4}\).

a) \(\frac{3}{4}\) and \(\frac{1}{2}\) b) d), because \(\frac{1}{2}=\frac{2}{4}\); each half is split into two fourths. c) \(\frac{1}{4}\)

- A \(6\times4\) grid can show sixths with full rows and fourths with full columns. - Inspect which candidates preserve the same portion of the whole as each original bar. - Use the \(24\) equal cells as the common unit before subtracting.

1. Panel a) shows \(\frac{5}{6}\), and panel b) shows \(\frac{1}{4}\). 2. Panel c) shades \(5\) full rows out of \(6\), so it represents \(\frac{5}{6}=\frac{20}{24}\). Panel f) shades \(1\) full column out of \(4\), so it represents \(\frac{1}{4}=\frac{6}{24}\). 3. The difference is \(\frac{20}{24}-\frac{6}{24}=\frac{14}{24}=\frac{7}{12}\).

a) \(\frac{5}{6}\) and \(\frac{1}{4}\) b) c) matches panel a) because it shades \(5\) of \(6\) full rows. f) matches panel b) because it shades \(1\) of \(4\) full columns. c) \(\frac{5}{6}=\frac{20}{24}\), \(\frac{1}{4}=\frac{6}{24}\), and the difference is \(\frac{7}{12}\).

- Read the shaded amount by counting complete shaded bars and the shaded part of the last bar. - Compare the fractional parts before subtracting. - If the top fractional part is too small, trade one whole for fourths.

1. The model shows \(3\frac{1}{4}\): three whole bars and one fourth of another bar are shaded. 2. Regroup one whole as \(\frac{4}{4}\), so \(3\frac{1}{4}=2\frac{5}{4}\). 3. Subtract: \(2\frac{5}{4}-1\frac{3}{4}=1\frac{2}{4}=1\frac{1}{2}\). 4. Regrouping keeps the value the same because one whole is equal to \(\frac{4}{4}\).

\(3\frac{1}{4}-1\frac{3}{4}=1\frac{1}{2}\). Regrouping one whole gives \(3\frac{1}{4}=2\frac{5}{4}\).

- After rewriting the mixed numbers as improper fractions, find a common denominator. - In part b), what can you do when the first fractional part is smaller than the second? - Compare the number and type of steps in the two methods.

1. For a), rewrite the mixed numbers: \(5 \frac{1}{6}=\frac{31}{6}\) and \(2 \frac{3}{4}=\frac{11}{4}\). 2. Use denominator \(12\): \(\frac{31}{6}-\frac{11}{4}=\frac{62}{12}-\frac{33}{12}=\frac{29}{12}=2 \frac{5}{12}\). 3. For b), regroup \(5 \frac{1}{6}\) as \(4 \frac{7}{6}\). 4. Use denominator \(12\) for the fractional parts: \(4 \frac{14}{12}-2 \frac{9}{12}=2 \frac{5}{12}\). Both methods give the same result; the preferred method can depend on which steps you find easier to track accurately.

a) \(2 \frac{5}{12}\) b) \(2 \frac{5}{12}\) Both methods are correct. Which one is less error-prone depends on the student's reasoning and organization.

- Look closely at the order of the numbers in \(2-5\). - Is the first fractional part large enough to subtract \(\frac{5}{8}\) from it directly? - How can one whole be regrouped as eighths?

1. The error occurs in the last step, where Selam changes \((8-3)+\frac{2-5}{8}\) into \(5 \frac{3}{8}\). 2. The fractional subtraction \(\frac{2}{8}-\frac{5}{8}\) cannot become positive \(\frac{3}{8}\). Because \(\frac{2}{8}<\frac{5}{8}\), one whole must be regrouped before subtracting the fractional parts. 3. Rewrite \(8 \frac{2}{8}\) as \(7 \frac{10}{8}\). Then \(7 \frac{10}{8}-3 \frac{5}{8}=4 \frac{5}{8}\).

1. The error occurs in the last step. 2. Selam treats \(\frac{2}{8}-\frac{5}{8}\) as positive \(\frac{3}{8}\), even though the fractional difference is negative before regrouping. 3. After regrouping one whole, the correct result is \(4 \frac{5}{8}\).

- Subtract \(\frac{1}{6}\) from both sides. - Find a common denominator for \(\frac{1}{5}\) and \(\frac{1}{6}\). - Match the resulting unit fraction to \(\frac{1}{x}\).

1. Isolate the unknown fraction: \(\frac{1}{x}=\frac{1}{5}-\frac{1}{6}\). 2. Subtract using denominator \(30\): \(\frac{1}{5}-\frac{1}{6}=\frac{6}{30}-\frac{5}{30}=\frac{1}{30}\). Therefore, \(x=30\). 3. Check: \(\frac{1}{6}+\frac{1}{30}=\frac{5}{30}+\frac{1}{30}=\frac{6}{30}=\frac{1}{5}\).

\(x=30\)

- Calculate each difference separately. - Then rewrite the two results in a form that makes them easy to compare.

1. Calculate \(A\): \(\frac{5}{2}-\frac{3}{4}=\frac{10}{4}-\frac{3}{4}=\frac{7}{4}\). 2. Calculate \(B\): \(\frac{11}{6}-\frac{1}{3}=\frac{11}{6}-\frac{2}{6}=\frac{9}{6}=\frac{3}{2}\). 3. Rewrite \(\frac{3}{2}\) as \(\frac{6}{4}\). Since \(\frac{7}{4}>\frac{6}{4}\), \(A>B\).

\(A>B\), because \(\frac{7}{4}>\frac{3}{2}\).

- Think about what quantity must be removed from the first fraction to leave the second. - Rewrite the two known fractions using equal-sized parts. - Check the missing value by placing it back into the original equation.

1. Rearrange the relationship as \(x=\frac{7}{9}-\frac{5}{12}\). 2. Use denominator \(36\): \(\frac{28}{36}-\frac{15}{36}=\frac{13}{36}\). 3. Substituting \(\frac{13}{36}\) gives \(\frac{28}{36}-\frac{13}{36}=\frac{15}{36}=\frac{5}{12}\).

\(x=\frac{13}{36}\).

- Rewrite both fractions so their parts are the same size. - After subtracting, compare the result with an equivalent form of the benchmark fraction.

1. Use denominator \(36\): \(\frac{11}{12}=\frac{33}{36}\) and \(\frac{5}{18}=\frac{10}{36}\). 2. Subtract: \(\frac{33}{36}-\frac{10}{36}=\frac{23}{36}\). 3. Since \(\frac{2}{3}=\frac{24}{36}\), the difference is less than \(\frac{2}{3}\).

The difference is \(\frac{23}{36}\), which is less than \(\frac{2}{3}\).

- When the starting value is the same, compare how much is being taken away. - Then use common denominators to verify the exact differences.

1. Both expressions start with \(\frac{5}{6}\). Subtracting \(\frac{1}{5}\) removes less than subtracting \(\frac{1}{4}\), so the first difference is greater. 2. \(\frac{5}{6}-\frac{1}{5}=\frac{25}{30}-\frac{6}{30}=\frac{19}{30}\). 3. \(\frac{5}{6}-\frac{1}{4}=\frac{10}{12}-\frac{3}{12}=\frac{7}{12}\), and \(\frac{19}{30}>\frac{7}{12}\).

\(\frac{5}{6}-\frac{1}{5}\) is greater; the differences are \(\frac{19}{30}\) and \(\frac{7}{12}\).

- Check whether the original fractional parts have the same size. - Think about what must be true of the denominators before the numerators can be subtracted. - After renaming the fractions, compare your result with the size of the original fractions.

1. Lucía subtracted the numerators and denominators separately, but fourths and sixths are different-sized parts, so they must first be renamed with a common denominator. 2. Rewrite with denominator \(12\): \(\frac{3}{4}=\frac{9}{12}\) and \(\frac{1}{6}=\frac{2}{12}\). 3. Subtract: \(\frac{9}{12}-\frac{2}{12}=\frac{7}{12}\).

Lucía subtracted the numerators and denominators separately even though the fractions name different-sized parts. After using a common denominator, the correct difference is \(\frac{7}{12}\).

- Rewrite the whole number as a fraction for the first difference. - Use a common denominator for the second difference. - Compare the two results using the same denominator.

1. \(A=\frac{12}{12}-\frac{5}{12}=\frac{7}{12}=\frac{14}{24}\). 2. \(B=\frac{21}{24}-\frac{8}{24}=\frac{13}{24}\). 3. \(A-B=\frac{14}{24}-\frac{13}{24}=\frac{1}{24}\).

\(A\) is greater than \(B\) by \(\frac{1}{24}\).

- Think about which operation undoes adding the known fraction. - Rewrite the quantities with a common denominator before subtracting. - Check your result by adding it to the known addend.

1. The missing addend is \(x=2\frac{1}{5}-\frac{7}{8}\). 2. Rewrite \(2\frac{1}{5}=\frac{11}{5}=\frac{88}{40}\) and \(\frac{7}{8}=\frac{35}{40}\). 3. Subtract: \(x=\frac{88}{40}-\frac{35}{40}=\frac{53}{40}=1\frac{13}{40}\).

\(x=1\frac{13}{40}\)

- Converting to improper fractions can handle the needed regrouping automatically. - Use a common denominator before subtracting. - Convert the improper result back to a mixed number.

1. Rewrite the mixed numbers as improper fractions: \(3\frac{1}{4}=\frac{13}{4}\) and \(1\frac{5}{6}=\frac{11}{6}\). 2. Use denominator \(12\): \(\frac{39}{12}-\frac{22}{12}=\frac{17}{12}\). 3. \(\frac{17}{12}=1\frac{5}{12}\).

\(1\frac{5}{12}\)

- Think about a common denominator that works for sixths and eighths. - Compare the exact difference with a familiar benchmark fraction after you subtract.

1. Rewrite \(1\frac{1}{6}=\frac{7}{6}\). 2. Use denominator \(24\): \(\frac{7}{6}=\frac{28}{24}\) and \(\frac{5}{8}=\frac{15}{24}\). 3. Subtract: \(\frac{28}{24}-\frac{15}{24}=\frac{13}{24}\). 4. Since \(\frac{13}{24}\) is just greater than \(\frac{12}{24}=\frac{1}{2}\), the estimate is reasonable.

The exact difference is \(\frac{13}{24}\), so the estimate is reasonable.

- Compare the two fractional parts before subtracting them. - Think about rewriting one whole as an equivalent fractional amount. - After regrouping, use equivalent fractions with a common denominator.

1. Since \(\frac{1}{4}<\frac{5}{6}\), the fractional parts cannot be subtracted directly while keeping the whole-number difference unchanged. 2. Rewrite \(2\frac{1}{4}=1\frac{5}{4}\). 3. Use twelfths: \(\frac{5}{4}=\frac{15}{12}\) and \(\frac{5}{6}=\frac{10}{12}\). 4. Subtract: \(1\frac{15}{12}-1\frac{10}{12}=\frac{5}{12}\).

Regrouping is needed because \(\frac{1}{4}<\frac{5}{6}\). The correct difference is \(\frac{5}{12}\).

- Find each difference using equivalent fractions with like denominators. - Before ordering, rewrite the final differences with comparable denominators if needed. - Be alert to the possibility that two different-looking expressions have the same value.

1. \(A=\frac{9}{12}-\frac{2}{12}=\frac{7}{12}\). 2. \(B=\frac{10}{12}-\frac{3}{12}=\frac{7}{12}\). 3. \(C=\frac{21}{24}-\frac{8}{24}=\frac{13}{24}\). 4. Rewrite \(\frac{7}{12}=\frac{14}{24}\). Since \(\frac{13}{24}<\frac{14}{24}\), the order is \(C<A=B\).

\(C<A=B\). Specifically, \(C=\frac{13}{24}\) and \(A=B=\frac{7}{12}\).

- Each subtraction involves fractions that are close together. - Find a common denominator for each pair and look at the gap between the equivalent numerators. - Compare the exact differences after simplifying.

1. \(A=\frac{21}{30}-\frac{20}{30}=\frac{1}{30}\). 2. \(B=\frac{25}{30}-\frac{24}{30}=\frac{1}{30}\). 3. The two differences are equal, so they are equally close to \(0\).

\(A=B=\frac{1}{30}\), so they are equally close to \(0\).

- Rewrite the subtrahend in each expression with the same denominator as the minuend. - Compare each simplified difference with an equivalent form of \(\frac{1}{2}\). - Do not assume expressions with similar-looking numbers have the same value.

1. a) \(\frac{5}{6}-\frac{1}{3}=\frac{5}{6}-\frac{2}{6}=\frac{3}{6}=\frac{1}{2}\). 2. b) \(\frac{7}{10}-\frac{1}{5}=\frac{7}{10}-\frac{2}{10}=\frac{5}{10}=\frac{1}{2}\). 3. c) \(\frac{11}{12}-\frac{1}{3}=\frac{11}{12}-\frac{4}{12}=\frac{7}{12}\), so it is not \(\frac{1}{2}\). 4. Therefore a) and b) are the correct choices.

a) and b). For a), \(\frac{5}{6}-\frac{1}{3}=\frac{3}{6}=\frac{1}{2}\). For b), \(\frac{7}{10}-\frac{1}{5}=\frac{5}{10}=\frac{1}{2}\). Choice c) equals \(\frac{7}{12}\), not \(\frac{1}{2}\).

- Rewrite the fractional parts with a common denominator. - Check whether the fractional part of the first mixed number is large enough to subtract the second fractional part. - Regroup one whole if needed, then simplify the final fraction.

1. Rewrite the fractional parts with denominator \(12\): \(\frac{1}{3}=\frac{4}{12}\). 2. Since \(\frac{4}{12}<\frac{7}{12}\), regroup \(6\frac{4}{12}=5\frac{16}{12}\). 3. Subtract: \(5\frac{16}{12}-2\frac{7}{12}=3\frac{9}{12}\). 4. Simplify \(\frac{9}{12}=\frac{3}{4}\), giving \(3\frac{3}{4}\).

\(3\frac{3}{4}\).

- Compare the two fractional parts after they have a common denominator. - Think about whether a whole needs to be regrouped into tenths. - Subtraction is directional; do not replace it with an absolute difference.

1. After rewriting, the fractional part \(\frac{2}{10}\) is smaller than \(\frac{7}{10}\), so Mateo cannot take an absolute difference while keeping the whole \(1\). 2. Regroup \(1\frac{2}{10}=\frac{12}{10}\). 3. Subtract: \(\frac{12}{10}-\frac{7}{10}=\frac{5}{10}=\frac{1}{2}\).

Mateo failed to regroup and incorrectly used an absolute difference. The correct difference is \(\frac{1}{2}\).

- Rewrite each subtraction so the fractions being subtracted have like denominators. - Keep each difference separate until both are simplified. - Compare the two final values and subtract them to find the gap.

1. \(A=\frac{24}{12}-\frac{7}{12}=\frac{17}{12}=1\frac{5}{12}\). 2. \(B=\frac{4}{3}-\frac{1}{4}=\frac{16}{12}-\frac{3}{12}=\frac{13}{12}=1\frac{1}{12}\). 3. The difference between the results is \(\frac{17}{12}-\frac{13}{12}=\frac{4}{12}=\frac{1}{3}\). 4. Therefore \(A\) is greater by \(\frac{1}{3}\).

\(A=1\frac{5}{12}\), \(B=1\frac{1}{12}\), and \(A\) is greater by \(\frac{1}{3}\).

- Use equivalent fractions so the subtraction and both comparisons use the same denominator. - After finding the difference, measure its distance from each benchmark. - The smaller distance identifies the closer benchmark.

1. Use denominator \(24\): \(\frac{17}{12}=\frac{34}{24}\) and \(\frac{5}{8}=\frac{15}{24}\). 2. The difference is \(\frac{19}{24}\). 3. Its distance from \(\frac{3}{4}=\frac{18}{24}\) is \(\frac{1}{24}\), while its distance from \(1=\frac{24}{24}\) is \(\frac{5}{24}\). 4. Therefore, the difference is closer to \(\frac{3}{4}\).

The difference is \(\frac{19}{24}\), and it is closer to \(\frac{3}{4}\).

- Use equivalent fractions so the subtraction and benchmark comparison share a denominator. - Find the exact difference before deciding its position relative to the benchmark. - Subtract the benchmark from the result to measure the gap.

1. Use denominator \(36\): \(\frac{13}{18}=\frac{26}{36}\) and \(\frac{5}{12}=\frac{15}{36}\). 2. The difference is \(\frac{11}{36}\). 3. Since \(\frac{1}{4}=\frac{9}{36}\), the difference is greater than \(\frac{1}{4}\) by \(\frac{2}{36}=\frac{1}{18}\).

The difference is \(\frac{11}{36}\), which is \(\frac{1}{18}\) greater than \(\frac{1}{4}\).

- Find a common denominator for ninths and twelfths. - Compare the fractional parts after rewriting them. - Regroup one whole if the first fractional part is too small to subtract the second.

1. Rewrite the fractional parts with denominator \(36\): \(\frac{5}{9}=\frac{20}{36}\) and \(\frac{7}{12}=\frac{21}{36}\). 2. Since \(\frac{20}{36}<\frac{21}{36}\), regroup \(2\frac{20}{36}=1\frac{56}{36}\). 3. Subtract: \(1\frac{56}{36}-1\frac{21}{36}=\frac{35}{36}\). 4. The result is already in simplest form.

\(\frac{35}{36}\).

- Notice whether the fractional part of the first mixed number is large enough to subtract the second fractional part. - One whole can be rewritten as a fraction with the same denominator as the fractional part. - After regrouping, use equivalent fractions with a common denominator.

1. Regroup \(5\frac{1}{8}\) as \(4\frac{9}{8}\). 2. Use denominator \(24\): \(\frac{9}{8}=\frac{27}{24}\) and \(\frac{5}{6}=\frac{20}{24}\). 3. Subtract: \(4\frac{27}{24}-2\frac{20}{24}=2\frac{7}{24}\).

Regroup \(5\frac{1}{8}\) as \(4\frac{9}{8}\). The difference is \(2\frac{7}{24}\).

- Compare each fraction with a familiar benchmark before doing exact arithmetic. - Use a common denominator for the exact subtraction. - Compare the result with your benchmark estimate to assess reasonableness.

1. \(\frac{11}{12}\) is close to \(1\), and \(\frac{2}{5}\) is close to \(\frac{1}{2}\), so a difference near \(\frac{1}{2}\) is reasonable. 2. Use denominator \(60\): \(\frac{11}{12}=\frac{55}{60}\) and \(\frac{2}{5}=\frac{24}{60}\). 3. Subtract: \(\frac{55}{60}-\frac{24}{60}=\frac{31}{60}\). 4. Since \(\frac{31}{60}\) is just greater than \(\frac{30}{60}=\frac{1}{2}\), the estimate was reasonable.

The exact difference is \(\frac{31}{60}\), which is close to \(\frac{1}{2}\).

- Rewrite both fractions with a denominator that also makes the benchmark easy to compare. - Find the difference first. - Compare the numerator of the result with the numerator representing one half.

1. Compare the expression directly with \(\frac{1}{2}\): \(\frac{7}{8}-\frac{1}{2}=\frac{3}{8}=\frac{21}{56}\). 2. Also, \(\frac{5}{14}=\frac{20}{56}\), so subtracting \(\frac{5}{14}\) removes \(\frac{1}{56}\) less than the amount that would bring \(\frac{7}{8}\) down to \(\frac{1}{2}\). 3. Therefore the difference is \(\frac{1}{56}\) above \(\frac{1}{2}\), so it equals \(\frac{29}{56}\).

The difference is \(\frac{29}{56}\), which is \(\frac{1}{56}\) above \(\frac{1}{2}\).

- Use one denominator that works for the fractional parts and for the benchmark. - Find the exact difference before making the final comparison. - Subtract the two nearby values to find how far apart they are.

1. Subtract using denominator \(60\): \(\frac{7}{10}=\frac{42}{60}\) and \(\frac{5}{12}=\frac{25}{60}\). 2. The difference is \(1\frac{17}{60}=\frac{77}{60}\). 3. Since \(1\frac{1}{4}=\frac{75}{60}\), the difference is greater by \(\frac{2}{60}=\frac{1}{30}\).

The difference is \(1\frac{17}{60}\), which is \(\frac{1}{30}\) greater than \(1\frac{1}{4}\).

- Think about which operation undoes the addition in the equation. - Keep the whole-number and fractional parts organized as you compare equivalent fractions. - Check the result by adding it back to the known addend.

1. Subtract \(1\frac{11}{15}\) from \(3\frac{7}{8}\): \(x=3\frac{7}{8}-1\frac{11}{15}\). 2. Use denominator \(120\): \(\frac{7}{8}=\frac{105}{120}\) and \(\frac{11}{15}=\frac{88}{120}\). 3. Subtract to get \(x=2\frac{17}{120}\).

\(x=2\frac{17}{120}\)

- Compare the two fractional parts before doing exact arithmetic. - Use equivalent fractions to make the exact subtraction possible. - Check whether the fractional part that remains is positive.

1. Because \(\frac{11}{12}>\frac{7}{15}\), subtracting \(\frac{7}{15}\) from the fractional part still leaves a positive fraction, so the result is greater than \(1\). 2. Use denominator \(60\): \(\frac{11}{12}=\frac{55}{60}\) and \(\frac{7}{15}=\frac{28}{60}\). 3. Subtract: \(1\frac{55}{60}-\frac{28}{60}=1\frac{27}{60}=1\frac{9}{20}\).

It is greater than \(1\). The exact difference is \(1\frac{9}{20}\).

- Read the original fractions from shaded parts over total parts in a) and b). - Count all cells in either common grid to identify the common unit. - Compare the shaded-cell counts only after both fractions are shown with that same unit.

1. Panel a) shows \(\frac{7}{8}\), and panel b) shows \(\frac{2}{3}\). 2. Each common grid has \(24\) equal cells, so one cell represents \(\frac{1}{24}\) of a whole. 3. Panel c) shows \(21\) shaded cells, so \(\frac{7}{8}=\frac{21}{24}\). Panel d) shows \(16\) shaded cells, so \(\frac{2}{3}=\frac{16}{24}\). 4. Eighths and thirds are different-sized parts, so their original numerator counts cannot be subtracted directly. With common units, the difference is \(\frac{21}{24}-\frac{16}{24}=\frac{5}{24}\).

a) \(\frac{7}{8}\) and \(\frac{2}{3}\) b) Each common-grid cell is \(\frac{1}{24}\). c) The original parts are different sizes. The grids show \(\frac{21}{24}-\frac{16}{24}=\frac{5}{24}\).