Question

rename and subtract the numbers. simplify your answers when necessary.
1.
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$$\begin{array}{r} 7 \\frac{1}{4} = 7 \\frac{}{8} = 6 - \\\\ -\\ \\ 5 \\frac{3}{8} = 5 \\frac{}{8} = 5 - \\\\ \\hline \\end{array}$$

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2.
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$$\begin{array}{r} 5 \\frac{1}{3} = \\\\ -\\ \\ 2 \\frac{5}{6} = \\\\ \\hline \\end{array}$$

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3.
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$$\begin{array}{r} 8 \\frac{7}{10} = \\\\ -\\ \\ 5 \\frac{4}{5} = \\\\ \\hline \\end{array}$$

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4.
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$$\begin{array}{r} 4 \\frac{7}{12} = \\\\ -\\ \\ 2 \\frac{3}{4} = \\\\ \\hline \\end{array}$$

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5.
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$$\begin{array}{r} 13 \\frac{1}{3} = 13 \\frac{}{9} = \\\\ -\\ \\ 9 \\frac{7}{9} = 9 \\frac{}{9} = \\\\ \\hline \\end{array}$$

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6.
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$$\begin{array}{r} 8 \\frac{1}{9} = \\\\ -\\ \\ 3 \\frac{5}{6} = \\\\ \\hline \\end{array}$$

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7.
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$$\begin{array}{r} 9 \\frac{3}{10} = \\\\ -\\ \\ 2 \\frac{1}{2} = \\\\ \\hline \\end{array}$$

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8.
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$$\begin{array}{r} 15 \\frac{1}{3} = \\\\ -\\ \\ 7 \\frac{1}{2} = \\\\ \\hline \\end{array}$$

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Explanation:

Step1: Rename \( 7\frac{1}{4} \) to eighths

To subtract mixed numbers, we need a common denominator. The denominator of the fraction in \( 5\frac{3}{8} \) is 8, so we convert \( \frac{1}{4} \) to eighths. Since \( \frac{1}{4}=\frac{1\times2}{4\times2}=\frac{2}{8} \), but we also need to borrow 1 from the whole number because \( \frac{2}{8}<\frac{3}{8} \). So \( 7\frac{1}{4}=6 + 1+\frac{1}{4}=6+\frac{4}{4}+\frac{1}{4}=6\frac{5}{4} \)? Wait, no, better to borrow 1 as \( \frac{8}{8} \). So \( 7\frac{1}{4}=7\frac{2}{8}=6 + 1\frac{2}{8}=6\frac{10}{8} \) (because \( 1=\frac{8}{8} \), so \( 7\frac{2}{8}=6 + 1\frac{2}{8}=6+\frac{8 + 2}{8}=6\frac{10}{8} \)). And \( 5\frac{3}{8} \) remains \( 5\frac{3}{8} \), but we can also write it as \( 5\frac{3}{8} \), and when subtracting, we have \( 6\frac{10}{8}-5\frac{3}{8} \).

Step2: Subtract the whole numbers and fractions

Subtract the whole numbers: \( 6 - 5 = 1 \). Subtract the fractions: \( \frac{10}{8}-\frac{3}{8}=\frac{7}{8} \). So the result is \( 1\frac{7}{8} \)? Wait, no, let's do it step by step as per the problem's renaming. The problem has \( 7\frac{1}{4}=7\frac{?}{8}=6 - \), wait, maybe the problem wants to rename \( 7\frac{1}{4} \) to a fraction with denominator 8 and also borrow 1 to make the fraction part larger. So \( 7\frac{1}{4}=7\frac{2}{8} \), but since \( \frac{2}{8}<\frac{3}{8} \), we borrow 1 from 7, making it 6, and add \( \frac{8}{8} \) to \( \frac{2}{8} \), so \( 7\frac{2}{8}=6\frac{10}{8} \). Then \( 5\frac{3}{8} \) is \( 5\frac{3}{8} \). Now subtract: \( 6\frac{10}{8}-5\frac{3}{8}=(6 - 5)+(\frac{10}{8}-\frac{3}{8})=1\frac{7}{8} \). Wait, but the problem's format is \( 7\frac{1}{4}=7\frac{?}{8}=6 - \), maybe they want \( 7\frac{1}{4}=7\frac{2}{8}=6 + 1\frac{2}{8}=6\frac{10}{8} \), so \( 7\frac{1}{4}=7\frac{2}{8}=6\frac{10}{8} \), and \( 5\frac{3}{8}=5\frac{3}{8} \). Then subtracting: \( 6\frac{10}{8}-5\frac{3}{8}=1\frac{7}{8} \).

Wait, maybe the first problem (1) is:
\( 7\frac{1}{4}-5\frac{3}{8} \)
Rename \( 7\frac{1}{4} \) to eighths: \( 7\frac{1}{4}=7\frac{2}{8} \), but we need to borrow 1 because \( \frac{2}{8}<\frac{3}{8} \), so \( 7\frac{2}{8}=6\frac{10}{8} \)
\( 5\frac{3}{8} \) remains \( 5\frac{3}{8} \)
Now subtract: \( 6\frac{10}{8}-5\frac{3}{8}=(6 - 5)+(\frac{10}{8}-\frac{3}{8})=1\frac{7}{8} \)

Let's do problem 2: \( 5\frac{1}{3}-2\frac{5}{6} \)
Rename \( 5\frac{1}{3} \) to sixths: \( \frac{1}{3}=\frac{2}{6} \), so \( 5\frac{1}{3}=5\frac{2}{6} \), but \( \frac{2}{6}<\frac{5}{6} \), so borrow 1: \( 5\frac{2}{6}=4\frac{8}{6} \)
Subtract \( 2\frac{5}{6} \) from \( 4\frac{8}{6} \): whole numbers \( 4 - 2 = 2 \), fractions \( \frac{8}{6}-\frac{5}{6}=\frac{3}{6}=\frac{1}{2} \), so result \( 2\frac{1}{2} \)

Problem 3: \( 8\frac{7}{10}-5\frac{4}{5} \)
Rename \( \frac{4}{5} \) to tenths: \( \frac{4}{5}=\frac{8}{10} \), so \( 5\frac{4}{5}=5\frac{8}{10} \)
\( 8\frac{7}{10} \) has \( \frac{7}{10}<\frac{8}{10} \), so borrow 1: \( 8\frac{7}{10}=7\frac{17}{10} \)
Subtract: \( 7\frac{17}{10}-5\frac{8}{10}=(7 - 5)+(\frac{17}{10}-\frac{8}{10})=2\frac{9}{10} \)

Problem 4: \( 4\frac{7}{12}-2\frac{3}{4} \)
Rename \( \frac{3}{4} \) to twelfths: \( \frac{3}{4}=\frac{9}{12} \), so \( 2\frac{3}{4}=2\frac{9}{12} \)
\( 4\frac{7}{12} \) has \( \frac{7}{12}<\frac{9}{12} \), so borrow 1: \( 4\frac{7}{12}=3\frac{19}{12} \)
Subtract: \( 3\frac{19}{12}-2\frac{9}{12}=(3 - 2)+(\frac{19}{12}-\frac{9}{12})=1\frac{10}{12}=1\frac{5}{6} \)

Problem 5: \( 13\frac{1}{3}-9\frac{7}{9} \)
Rename \( \frac{1}{3} \) to ninths: \( \frac{1}{3}=\frac{3}{9} \), so \( 13\frac{1}{3}=13\frac{3}{9} \), borrow 1: \( 13\frac{3}{9}=12\frac{12}{9} \)
Subtract \( 9\frac{7}{9} \): \( 12 - 9 = 3 \), \( \frac{12}{9}-\frac{7}{9}=\frac{5}{9} \), so \( 3\frac{5}{9} \)

Problem 6: \( 8\frac{1}{9}-3\frac{5}{6} \)
Find common denominator, which is 18. \( \frac{1}{9}=\frac{2}{18} \), \( \frac{5}{6}=\frac{15}{18} \)
\( 8\frac{2}{18} \) borrows 1: \( 8\frac{2}{18}=7\frac{20}{18} \)
Subtract \( 3\frac{15}{18} \): \( 7 - 3 = 4 \), \( \frac{20}{18}-\frac{15}{18}=\frac{5}{18} \), so \( 4\frac{5}{18} \)

Problem 7: \( 9\frac{3}{10}-2\frac{1}{2} \)
Rename \( \frac{1}{2} \) to tenths: \( \frac{1}{2}=\frac{5}{10} \), so \( 2\frac{1}{2}=2\frac{5}{10} \)
\( 9\frac{3}{10} \) has \( \frac{3}{10}<\frac{5}{10} \), borrow 1: \( 9\frac{3}{10}=8\frac{13}{10} \)
Subtract: \( 8 - 2 = 6 \), \( \frac{13}{10}-\frac{5}{10}=\frac{8}{10}=\frac{4}{5} \), so \( 6\frac{4}{5} \)

Problem 8: \( 15\frac{1}{3}-7\frac{1}{2} \)
Common denominator 6: \( \frac{1}{3}=\frac{2}{6} \), \( \frac{1}{2}=\frac{3}{6} \)
\( 15\frac{2}{6} \) borrows 1: \( 15\frac{2}{6}=14\frac{8}{6} \)
Subtract \( 7\frac{3}{6} \): \( 14 - 7 = 7 \), \( \frac{8}{6}-\frac{3}{6}=\frac{5}{6} \), so \( 7\frac{5}{6} \)

But let's focus on problem 1 as an example:

Step1: Rename \( 7\frac{1}{4} \) to eighths and borrow

To subtract \( 7\frac{1}{4}-5\frac{3}{8} \), first find a common denominator for the fractions. The denominators are 4 and 8, so the common denominator is 8. Convert \( \frac{1}{4} \) to eighths: \( \frac{1}{4}=\frac{1\times2}{4\times2}=\frac{2}{8} \). Now, since \( \frac{2}{8}<\frac{3}{8} \) (the fraction part of the subtrahend), we need to borrow 1 from the whole number part of the minuend. So, \( 7\frac{2}{8}=6 + 1\frac{2}{8}=6+\frac{8}{8}+\frac{2}{8}=6\frac{10}{8} \). The subtrahend \( 5\frac{3}{8} \) remains \( 5\frac{3}{8} \), and we can also write it as \( 5\frac{3}{8} \) (no need to borrow here).

Step2: Subtract the whole numbers and fractions

Now, subtract the whole number parts: \( 6 - 5 = 1 \). Subtract the fraction parts: \( \frac{10}{8}-\frac{3}{8}=\frac{7}{8} \). Combine the results: \( 1+\frac{7}{8}=1\frac{7}{8} \).

For the renaming part as per the problem's format:
\( 7\frac{1}{4}=7\frac{2}{8}=6\frac{10}{8} \) (because we borrowed 1, so 7 becomes 6 and we add \( \frac{8}{8} \) to \( \frac{2}{8} \))
\( 5\frac{3}{8}=5\frac{3}{8}=5\frac{3}{8} \) (but in the problem's blanks, maybe they want \( 7\frac{1}{4}=7\frac{2}{8}=6\frac{10}{8} \) and \( 5\frac{3}{8}=5\frac{3}{8}=5\frac{3}{8} \), then subtracting \( 6\frac{10}{8}-5\frac{3}{8}=1\frac{7}{8} \)

Answer:

1. \( 7\frac{1}{4}=7\frac{2}{8}=6\frac{10}{8} \); \( 5\frac{3}{8}=5\frac{3}{8}=5\frac{3}{8} \); Result: \( 1\frac{7}{8} \)
2. \( 5\frac{1}{3}=5\frac{2}{6}=4\frac{8}{6} \); \( 2\frac{5}{6}=2\frac{5}{6} \); Result: \( 2\frac{1}{2} \)
3. \( 8\frac{7}{10}=8\frac{7}{10} \); \( 5\frac{4}{5}=5\frac{8}{10} \); \( 8\frac{7}{10}=7\frac{17}{10} \); Result: \( 2\frac{9}{10} \)
4. \( 4\frac{7}{12}=4\frac{7}{12} \); \( 2\frac{3}{4}=2\frac{9}{12} \); \( 4\frac{7}{12}=3\frac{19}{12} \); Result: \( 1\frac{5}{6} \)
5. \( 13\frac{1}{3}=13\frac{3}{9}=12\frac{12}{9} \); \( 9\frac{7}{9}=9\frac{7}{9} \); Result: \( 3\frac{5}{9} \)
6. \( 8\frac{1}{9}=8\frac{2}{18}=7\frac{20}{18} \); \( 3\frac{5}{6}=3\frac{15}{18} \); Result: \( 4\frac{5}{18} \)
7. \( 9\frac{3}{10}=9\frac{3}{10} \); \( 2\frac{1}{2}=2\frac{5}{10} \); \( 9\frac{3}{10}=8\frac{13}{10} \); Result: \( 6\frac{4}{5} \)
8. \( 15\frac{1}{3}=15\frac{2}{6}=14\frac{8}{6} \); \( 7\frac{1}{2}=7\frac{3}{6} \); Result: \( 7\frac{5}{6} \)

(Note: The answers are provided for each problem as per the renaming and subtraction process. Each problem follows the steps of renaming the mixed numbers to have a common denominator, borrowing 1 from the whole number if necessary, and then subtracting the whole numbers and fractions separately, simplifying the result if needed.)