Math worksheet for subtracting mixed numbers with regrouping, including six practice problems.

Let's solve each of these subtracting mixed numbers with regrouping problems step by step. The key idea is that when the fractional part of the minuend (top number) is smaller than the fractional part of the subtrahend (bottom number), we need to regroup — borrow 1 from the whole number part and convert it into a fraction with the same denominator.

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🔷 Problem 1: $ 5\frac{1}{7} - 1\frac{3}{7} $

✔ This one is already solved for us, but let’s verify:

- Since $ \frac{1}{7} < \frac{3}{7} $, we need to borrow.
- Borrow 1 from 5 → becomes 4, and add $ \frac{7}{7} $ to $ \frac{1}{7} $ → $ \frac{8}{7} $
- So $ 5\frac{1}{7} = 4\frac{8}{7} $
- Now subtract:
$$
4\frac{8}{7} - 1\frac{3}{7} = (4 - 1) + \left(\frac{8}{7} - \frac{3}{7}\right) = 3 + \frac{5}{7} = \boxed{3\frac{5}{7}}
$$

✔️ Correct.

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🔷 Problem 2: $ 7\frac{1}{4} - 4\frac{2}{4} $

- $ \frac{1}{4} < \frac{2}{4} $ → need to borrow
- Borrow 1 from 7 → becomes 6, and add $ \frac{4}{4} $ to $ \frac{1}{4} $ → $ \frac{5}{4} $
- So $ 7\frac{1}{4} = 6\frac{5}{4} $
- Subtract:
$$
6\frac{5}{4} - 4\frac{2}{4} = (6 - 4) + \left(\frac{5}{4} - \frac{2}{4}\right) = 2 + \frac{3}{4} = \boxed{2\frac{3}{4}}
$$

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🔷 Problem 3: $ 8\frac{2}{5} - 2\frac{3}{5} $

- $ \frac{2}{5} < \frac{3}{5} $ → borrow
- Borrow 1 from 8 → 7, add $ \frac{5}{5} $ to $ \frac{2}{5} $ → $ \frac{7}{5} $
- So $ 8\frac{2}{5} = 7\frac{7}{5} $
- Subtract:
$$
7\frac{7}{5} - 2\frac{3}{5} = (7 - 2) + \left(\frac{7}{5} - \frac{3}{5}\right) = 5 + \frac{4}{5} = \boxed{5\frac{4}{5}}
$$

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🔷 Problem 4: $ 3\frac{4}{8} - 1\frac{7}{8} $

- Simplify first: $ \frac{4}{8} = \frac{1}{2} $, but keep as $ \frac{4}{8} $ for now
- $ \frac{4}{8} < \frac{7}{8} $ → borrow
- Borrow 1 from 3 → 2, add $ \frac{8}{8} $ to $ \frac{4}{8} $ → $ \frac{12}{8} $
- So $ 3\frac{4}{8} = 2\frac{12}{8} $
- Subtract:
$$
2\frac{12}{8} - 1\frac{7}{8} = (2 - 1) + \left(\frac{12}{8} - \frac{7}{8}\right) = 1 + \frac{5}{8} = \boxed{1\frac{5}{8}}
$$

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🔷 Problem 5: $ 9\frac{1}{9} - 4\frac{7}{9} $

- $ \frac{1}{9} < \frac{7}{9} $ → borrow
- Borrow 1 from 9 → 8, add $ \frac{9}{9} $ to $ \frac{1}{9} $ → $ \frac{10}{9} $
- So $ 9\frac{1}{9} = 8\frac{10}{9} $
- Subtract:
$$
8\frac{10}{9} - 4\frac{7}{9} = (8 - 4) + \left(\frac{10}{9} - \frac{7}{9}\right) = 4 + \frac{3}{9} = 4 + \frac{1}{3} = \boxed{4\frac{1}{3}}
$$

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🔷 Problem 6: $ 4\frac{1}{8} - 2\frac{2}{8} $

- $ \frac{1}{8} < \frac{2}{8} $ → borrow
- Borrow 1 from 4 → 3, add $ \frac{8}{8} $ to $ \frac{1}{8} $ → $ \frac{9}{8} $
- So $ 4\frac{1}{8} = 3\frac{9}{8} $
- Subtract:
$$
3\frac{9}{8} - 2\frac{2}{8} = (3 - 2) + \left(\frac{9}{8} - \frac{2}{8}\right) = 1 + \frac{7}{8} = \boxed{1\frac{7}{8}}
$$

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🔷 Problem 7: $ 6\frac{3}{6} - 1\frac{5}{6} $

- Simplify: $ \frac{3}{6} = \frac{1}{2} $, but keep as $ \frac{3}{6} $
- $ \frac{3}{6} < \frac{5}{6} $ → borrow
- Borrow 1 from 6 → 5, add $ \frac{6}{6} $ to $ \frac{3}{6} $ → $ \frac{9}{6} $
- So $ 6\frac{3}{6} = 5\frac{9}{6} $
- Subtract:
$$
5\frac{9}{6} - 1\frac{5}{6} = (5 - 1) + \left(\frac{9}{6} - \frac{5}{6}\right) = 4 + \frac{4}{6} = 4 + \frac{2}{3} = \boxed{4\frac{2}{3}}
$$

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🔷 Problem 8: $ 7\frac{1}{5} - 2\frac{2}{5} $

- $ \frac{1}{5} < \frac{2}{5} $ → borrow
- Borrow 1 from 7 → 6, add $ \frac{5}{5} $ to $ \frac{1}{5} $ → $ \frac{6}{5} $
- So $ 7\frac{1}{5} = 6\frac{6}{5} $
- Subtract:
$$
6\frac{6}{5} - 2\frac{2}{5} = (6 - 2) + \left(\frac{6}{5} - \frac{2}{5}\right) = 4 + \frac{4}{5} = \boxed{4\frac{4}{5}}
$$

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🔷 Problem 9: $ 8\frac{2}{4} - 4\frac{2}{4} $

- First simplify: $ \frac{2}{4} = \frac{1}{2} $
- But since both fractions are same: $ \frac{2}{4} $
- Whole numbers: $ 8 - 4 = 4 $
- Fractions: $ \frac{2}{4} - \frac{2}{4} = 0 $
- So result: $ \boxed{4} $

No regrouping needed here!

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🔷 Problem 10: $ 6\frac{3}{5} - 3\frac{4}{5} $

- $ \frac{3}{5} < \frac{4}{5} $ → borrow
- Borrow 1 from 6 → 5, add $ \frac{5}{5} $ to $ \frac{3}{5} $ → $ \frac{8}{5} $
- So $ 6\frac{3}{5} = 5\frac{8}{5} $
- Subtract:
$$
5\frac{8}{5} - 3\frac{4}{5} = (5 - 3) + \left(\frac{8}{5} - \frac{4}{5}\right) = 2 + \frac{4}{5} = \boxed{2\frac{4}{5}}
$$

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🔷 Problem 11: $ 4\frac{3}{8} - 1\frac{5}{8} $

- $ \frac{3}{8} < \frac{5}{8} $ → borrow
- Borrow 1 from 4 → 3, add $ \frac{8}{8} $ to $ \frac{3}{8} $ → $ \frac{11}{8} $
- So $ 4\frac{3}{8} = 3\frac{11}{8} $
- Subtract:
$$
3\frac{11}{8} - 1\frac{5}{8} = (3 - 1) + \left(\frac{11}{8} - \frac{5}{8}\right) = 2 + \frac{6}{8} = 2 + \frac{3}{4} = \boxed{2\frac{3}{4}}
$$

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🔷 Problem 12: $ 5\frac{9}{11} - 3\frac{10}{11} $

- $ \frac{9}{11} < \frac{10}{11} $ → borrow
- Borrow 1 from 5 → 4, add $ \frac{11}{11} $ to $ \frac{9}{11} $ → $ \frac{20}{11} $
- So $ 5\frac{9}{11} = 4\frac{20}{11} $
- Subtract:
$$
4\frac{20}{11} - 3\frac{10}{11} = (4 - 3) + \left(\frac{20}{11} - \frac{10}{11}\right) = 1 + \frac{10}{11} = \boxed{1\frac{10}{11}}
$$

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## ✔ Final Answers:

| Problem | Answer |
|--------|--------|
| 1 | $ 3\frac{5}{7} $ |
| 2 | $ 2\frac{3}{4} $ |
| 3 | $ 5\frac{4}{5} $ |
| 4 | $ 1\frac{5}{8} $ |
| 5 | $ 4\frac{1}{3} $ |
| 6 | $ 1\frac{7}{8} $ |
| 7 | $ 4\frac{2}{3} $ |
| 8 | $ 4\frac{4}{5} $ |
| 9 | $ 4 $ |
| 10 | $ 2\frac{4}{5} $ |
| 11 | $ 2\frac{3}{4} $ |
| 12 | $ 1\frac{10}{11} $ |

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📝 Summary of Method:

To subtract mixed numbers with regrouping:
1. Check if the top fraction is smaller than the bottom one.
2. If yes, borrow 1 from the whole number part.
3. Convert that borrowed 1 into a fraction with the same denominator.
4. Add it to the existing fraction.
5. Now subtract the whole numbers and the fractions separately.
6. Simplify if needed.

Let me know if you'd like this in printable format or with visual steps!