Question

rename the top mixed number and subtract. simplify when necessary.
1.
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$$\begin{array}{r} 7\\frac{3}{8} = \\\\ -\\ 4\\frac{5}{8} = \\\\ \\hline \\end{array}$$

4.
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$$\begin{array}{r} 16\\frac{4}{6} = \\\\ -\\ 8\\frac{5}{6} = \\\\ \\hline \\end{array}$$

7.
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$$\begin{array}{r} 3\\frac{1}{4} = \\\\ -\\ 1\\frac{3}{4} = \\\\ \\hline \\end{array}$$

2.
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$$\begin{array}{r} 6\\frac{2}{5} = \\\\ -\\ 1\\frac{4}{5} = \\\\ \\hline \\end{array}$$

5.
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$$\begin{array}{r} 7\\frac{3}{7} = \\\\ -\\ 5\\frac{6}{7} = \\\\ \\hline \\end{array}$$

8.
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$$\begin{array}{r} 7\\frac{2}{5} = \\\\ -\\ \\frac{3}{5} = \\\\ \\hline \\end{array}$$

3.
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$$\begin{array}{r} 20\\frac{1}{6} = \\\\ -\\ 15\\frac{5}{6} = \\\\ \\hline \\end{array}$$

6.
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$$\begin{array}{r} 4\\frac{4}{10} = \\\\ -\\ \\frac{7}{10} = \\\\ \\hline \\end{array}$$

9.
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$$\begin{array}{r} 10\\frac{4}{9} = \\\\ -\\ 6\\frac{7}{9} = \\\\ \\hline \\end{array}$$

Explanation:

Step1: Rename the top mixed number (Problem 4: \(16\frac{4}{6} - 8\frac{5}{6}\))

To subtract mixed numbers where the fraction part of the top number is less than the bottom, we rename the top mixed number. \(16\frac{4}{6}\) can be rewritten as \(15 + 1 + \frac{4}{6}\), and \(1 = \frac{6}{6}\), so \(16\frac{4}{6}=15\frac{6 + 4}{6}=15\frac{10}{6}\).

Step2: Subtract the whole numbers and fractions

Now subtract the whole numbers: \(15 - 8 = 7\), and subtract the fractions: \(\frac{10}{6}-\frac{5}{6}=\frac{10 - 5}{6}=\frac{5}{6}\). Combine them: \(7\frac{5}{6}\). Wait, but let's check again. Wait, \(16\frac{4}{6}=15\frac{10}{6}\), then \(15\frac{10}{6}-8\frac{5}{6}=(15 - 8)+(\frac{10}{6}-\frac{5}{6}) = 7+\frac{5}{6}=7\frac{5}{6}\)? Wait, no, wait \(16\frac{4}{6}\) is equal to \(15 + 1+\frac{4}{6}=15+\frac{6 + 4}{6}=15\frac{10}{6}\). Then subtracting \(8\frac{5}{6}\), so whole number part: \(15 - 8 = 7\), fraction part: \(\frac{10}{6}-\frac{5}{6}=\frac{5}{6}\), so \(7\frac{5}{6}\). But let's simplify \(\frac{5}{6}\) is already simplified. Wait, but maybe I made a mistake in renaming. Wait, \(16\frac{4}{6}\) can also be simplified first: \(16\frac{4}{6}=16\frac{2}{3}\), and \(8\frac{5}{6}\). Then convert to improper fractions: \(16\frac{2}{3}=\frac{16\times3 + 2}{3}=\frac{50}{3}\), \(8\frac{5}{6}=\frac{8\times6+5}{6}=\frac{53}{6}\). Find a common denominator, \(\frac{50}{3}=\frac{100}{6}\), then \(\frac{100}{6}-\frac{53}{6}=\frac{47}{6}=7\frac{5}{6}\). Yes, that's correct.

Let's do another problem (Problem 7: \(3\frac{1}{4}-1\frac{3}{4}\))

Step1: Rename the top mixed number

\(3\frac{1}{4}\) has a fraction part less than \(\frac{3}{4}\), so rename it as \(2 + 1+\frac{1}{4}\), \(1=\frac{4}{4}\), so \(3\frac{1}{4}=2\frac{4 + 1}{4}=2\frac{5}{4}\).

Step2: Subtract

Subtract whole numbers: \(2 - 1 = 1\), subtract fractions: \(\frac{5}{4}-\frac{3}{4}=\frac{2}{4}=\frac{1}{2}\). So the result is \(1\frac{1}{2}\).

Let's do Problem 3: \(20\frac{1}{6}-15\frac{5}{6}\)

Step1: Rename the top mixed number

\(20\frac{1}{6}\) becomes \(19\frac{6 + 1}{6}=19\frac{7}{6}\) (since \(20 = 19 + 1\) and \(1=\frac{6}{6}\), so \(20\frac{1}{6}=19+\frac{6}{6}+\frac{1}{6}=19\frac{7}{6}\)).

Step2: Subtract

Whole numbers: \(19 - 15 = 4\), fractions: \(\frac{7}{6}-\frac{5}{6}=\frac{2}{6}=\frac{1}{3}\). So \(4\frac{1}{3}\).

Let's do Problem 6: \(4\frac{4}{10}-\frac{7}{10}\)

Step1: Rename the top mixed number

\(4\frac{4}{10}\) has a fraction part less than \(\frac{7}{10}\), so rename it as \(3 + 1+\frac{4}{10}\), \(1=\frac{10}{10}\), so \(4\frac{4}{10}=3\frac{10 + 4}{10}=3\frac{14}{10}\).

Step2: Subtract

Whole number: \(3 - 0 = 3\) (since the second number has no whole number part), fractions: \(\frac{14}{10}-\frac{7}{10}=\frac{7}{10}\). Wait, no, the second number is \(\frac{7}{10}\), which is a proper fraction, so the whole number part of the top is 4, but we need to rename because \(\frac{4}{10}<\frac{7}{10}\). So \(4\frac{4}{10}=3\frac{14}{10}\), then \(3\frac{14}{10}-\frac{7}{10}=3+\frac{14 - 7}{10}=3\frac{7}{10}\).

Let's do Problem 9: \(10\frac{4}{9}-6\frac{7}{9}\)

Step1: Rename the top mixed number

\(10\frac{4}{9}\) becomes \(9\frac{9 + 4}{9}=9\frac{13}{9}\) (since \(10 = 9 + 1\) and \(1=\frac{9}{9}\)).

Step2: Subtract

Whole numbers: \(9 - 6 = 3\), fractions: \(\frac{13}{9}-\frac{7}{9}=\frac{6}{9}=\frac{2}{3}\). So \(3\frac{2}{3}\).

Let's redo Problem 1: \(7\frac{3}{8}-4\frac{5}{8}\)

Step1: Rename the top mixed number

\(7\frac{3}{8}\) has \(\frac{3}{8}<\frac{5}{8}\), so rename to \(6\frac{8 + 3}{8}=6\frac{11}{8}\).

Step2: Subtract

Whole numbers: \(6 - 4 = 2\), fractions: \(\frac{11}{8}-\frac{5}{8}=\frac{6}{8}=\frac{3}{4}\). So \(2\frac{3}{4}\) (the original handwritten answer was \(2\frac{6}{8}\) which simplifies to \(2\frac{3}{4}\)).

Problem 2: \(6\frac{2}{5}-1\frac{4}{5}\)

Step1: Rename the top mixed number

\(6\frac{2}{5}\) becomes \(5\frac{5 + 2}{5}=5\frac{7}{5}\).

Step2: Subtract

Whole numbers: \(5 - 1 = 4\), fractions: \(\frac{7}{5}-\frac{4}{5}=\frac{3}{5}\). So \(4\frac{3}{5}\) (the original handwritten answer was \(5\frac{3}{5}\) which is incorrect, correct is \(4\frac{3}{5}\)).

Problem 5: \(7\frac{3}{7}-5\frac{6}{7}\)

Step1: Rename the top mixed number

\(7\frac{3}{7}\) becomes \(6\frac{7 + 3}{7}=6\frac{10}{7}\).

Step2: Subtract

Whole numbers: \(6 - 5 = 1\), fractions: \(\frac{10}{7}-\frac{6}{7}=\frac{4}{7}\). So \(1\frac{4}{7}\) (the original handwritten answer was \(1\frac{4}{7}\), correct).

Problem 8: \(7\frac{2}{5}-\frac{3}{5}\)

Step1: Rename the top mixed number (if needed)

\(7\frac{2}{5}\) has \(\frac{2}{5}<\frac{3}{5}\)? Wait, no, \(\frac{2}{5}\) is less than \(\frac{3}{5}\), so rename \(7\frac{2}{5}\) to \(6\frac{5 + 2}{5}=6\frac{7}{5}\).

Step2: Subtract

Whole numbers: \(6 - 0 = 6\) (since the second number has no whole number), fractions: \(\frac{7}{5}-\frac{3}{5}=\frac{4}{5}\). So \(6\frac{4}{5}\) (the original handwritten answer was \(7\frac{2}{5}+\frac{8}{5}\)? Wait, no, correct process: \(7\frac{2}{5}=6\frac{7}{5}\), \(6\frac{7}{5}-\frac{3}{5}=6+\frac{4}{5}=6\frac{4}{5}\).

Now, let's present the solutions for each problem:

Problem 1: \(7\frac{3}{8}-4\frac{5}{8}\)

- Rename: \(7\frac{3}{8}=6\frac{11}{8}\)
- Subtract: \(6\frac{11}{8}-4\frac{5}{8}=(6 - 4)+(\frac{11}{8}-\frac{5}{8}) = 2+\frac{6}{8}=2\frac{3}{4}\)

Problem 2: \(6\frac{2}{5}-1\frac{4}{5}\)

- Rename: \(6\frac{2}{5}=5\frac{7}{5}\)
- Subtract: \(5\frac{7}{5}-1\frac{4}{5}=(5 - 1)+(\frac{7}{5}-\frac{4}{5}) = 4+\frac{3}{5}=4\frac{3}{5}\)

Problem 3: \(20\frac{1}{6}-15\frac{5}{6}\)

- Rename: \(20\frac{1}{6}=19\frac{7}{6}\)
- Subtract: \(19\frac{7}{6}-15\frac{5}{6}=(19 - 15)+(\frac{7}{6}-\frac{5}{6}) = 4+\frac{2}{6}=4\frac{1}{3}\)

Problem 4: \(16\frac{4}{6}-8\frac{5}{6}\)

- Rename: \(16\frac{4}{6}=15\frac{10}{6}\)
- Subtract: \(15\frac{10}{6}-8\frac{5}{6}=(15 - 8)+(\frac{10}{6}-\frac{5}{6}) = 7+\frac{5}{6}=7\frac{5}{6}\)

Problem 5: \(7\frac{3}{7}-5\frac{6}{7}\)

- Rename: \(7\frac{3}{7}=6\frac{10}{7}\)
- Subtract: \(6\frac{10}{7}-5\frac{6}{7}=(6 - 5)+(\frac{10}{7}-\frac{6}{7}) = 1+\frac{4}{7}=1\frac{4}{7}\)

Problem 6: \(4\frac{4}{10}-\frac{7}{10}\)

- Rename: \(4\frac{4}{10}=3\frac{14}{10}\)
- Subtract: \(3\frac{14}{10}-\frac{7}{10}=3+\frac{7}{10}=3\frac{7}{10}\)

Problem 7: \(3\frac{1}{4}-1\frac{3}{4}\)

- Rename: \(3\frac{1}{4}=2\frac{5}{4}\)
- Subtract: \(2\frac{5}{4}-1\frac{3}{4}=(2 - 1)+(\frac{5}{4}-\frac{3}{4}) = 1+\frac{2}{4}=1\frac{1}{2}\)

Problem 8: \(7\frac{2}{5}-\frac{3}{5}\)

- Rename: \(7\frac{2}{5}=6\frac{7}{5}\)
- Subtract: \(6\frac{7}{5}-\frac{3}{5}=6+\frac{4}{5}=6\frac{4}{5}\)

Problem 9: \(10\frac{4}{9}-6\frac{7}{9}\)

- Rename: \(10\frac{4}{9}=9\frac{13}{9}\)
- Subtract: \(9\frac{13}{9}-6\frac{7}{9}=(9 - 6)+(\frac{13}{9}-\frac{7}{9}) = 3+\frac{6}{9}=3\frac{2}{3}\)

Answers (for each problem):

1. \(2\frac{3}{4}\)
2. \(4\frac{3}{5}\)
3. \(4\frac{1}{3}\)
4. \(7\frac{5}{6}\)
5. \(1\frac{4}{7}\)
6. \(3\frac{7}{10}\)
7. \(1\frac{1}{2}\)
8. \(6\frac{4}{5}\)
9. \(3\frac{2}{3}\)

Answer:

s (for each problem):

1. \(2\frac{3}{4}\)
2. \(4\frac{3}{5}\)
3. \(4\frac{1}{3}\)
4. \(7\frac{5}{6}\)
5. \(1\frac{4}{7}\)
6. \(3\frac{7}{10}\)
7. \(1\frac{1}{2}\)
8. \(6\frac{4}{5}\)
9. \(3\frac{2}{3}\)