Adding Fractions with Same Denominator | Free Worksheets | Printables - Free Printable

Let's solve each of these fraction addition problems step by step. The key to adding fractions with unlike denominators is to:

1. Find the Least Common Denominator (LCD).
2. Convert each fraction to an equivalent fraction with the LCD as the denominator.
3. Add the numerators, keeping the denominator the same.
4. Simplify the result if needed.

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🔹 Problem 1: $ \frac{1}{4} + \frac{1}{2} $

- LCD of 4 and 2 is 4
- Convert $ \frac{1}{2} = \frac{2}{4} $
- $ \frac{1}{4} + \frac{2}{4} = \frac{3}{4} $

✔ Answer: $ \frac{3}{4} $

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🔹 Problem 2: $ \frac{3}{4} + \frac{3}{8} $

- LCD of 4 and 8 is 8
- $ \frac{3}{4} = \frac{6}{8} $
- $ \frac{6}{8} + \frac{3}{8} = \frac{9}{8} $
- Simplify: $ \frac{9}{8} = 1\frac{1}{8} $

✔ Answer: $ 1\frac{1}{8} $

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🔹 Problem 3: $ \frac{2}{4} + \frac{5}{6} $

- Simplify $ \frac{2}{4} = \frac{1}{2} $
- LCD of 2 and 6 is 6
- $ \frac{1}{2} = \frac{3}{6} $
- $ \frac{3}{6} + \frac{5}{6} = \frac{8}{6} $
- Simplify: $ \frac{8}{6} = \frac{4}{3} = 1\frac{1}{3} $

✔ Answer: $ 1\frac{1}{3} $

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🔹 Problem 4: $ \frac{3}{6} + \frac{1}{3} $

- Simplify $ \frac{3}{6} = \frac{1}{2} $
- LCD of 2 and 3 is 6
- $ \frac{1}{2} = \frac{3}{6}, \quad \frac{1}{3} = \frac{2}{6} $
- $ \frac{3}{6} + \frac{2}{6} = \frac{5}{6} $

✔ Answer: $ \frac{5}{6} $

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🔹 Problem 5: $ \frac{8}{3} + \frac{4}{5} $

- LCD of 3 and 5 is 15
- $ \frac{8}{3} = \frac{40}{15}, \quad \frac{4}{5} = \frac{12}{15} $
- $ \frac{40}{15} + \frac{12}{15} = \frac{52}{15} $
- Simplify: $ \frac{52}{15} = 3\frac{7}{15} $

✔ Answer: $ 3\frac{7}{15} $

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🔹 Problem 6: $ \frac{2}{5} + \frac{5}{10} $

- Simplify $ \frac{5}{10} = \frac{1}{2} $
- LCD of 5 and 2 is 10
- $ \frac{2}{5} = \frac{4}{10}, \quad \frac{1}{2} = \frac{5}{10} $
- $ \frac{4}{10} + \frac{5}{10} = \frac{9}{10} $

✔ Answer: $ \frac{9}{10} $

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🔹 Problem 7: $ \frac{2}{3} + \frac{1}{2} $

- LCD of 3 and 2 is 6
- $ \frac{2}{3} = \frac{4}{6}, \quad \frac{1}{2} = \frac{3}{6} $
- $ \frac{4}{6} + \frac{3}{6} = \frac{7}{6} $
- Simplify: $ \frac{7}{6} = 1\frac{1}{6} $

✔ Answer: $ 1\frac{1}{6} $

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🔹 Problem 8: $ \frac{4}{6} + \frac{5}{8} $

- Simplify $ \frac{4}{6} = \frac{2}{3} $
- LCD of 3 and 8 is 24
- $ \frac{2}{3} = \frac{16}{24}, \quad \frac{5}{8} = \frac{15}{24} $
- $ \frac{16}{24} + \frac{15}{24} = \frac{31}{24} $
- Simplify: $ \frac{31}{24} = 1\frac{7}{24} $

✔ Answer: $ 1\frac{7}{24} $

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🔹 Problem 9: $ \frac{3}{9} + \frac{1}{3} $

- Simplify $ \frac{3}{9} = \frac{1}{3} $
- $ \frac{1}{3} + \frac{1}{3} = \frac{2}{3} $

✔ Answer: $ \frac{2}{3} $

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🔹 Problem 10: $ \frac{6}{4} + \frac{5}{12} $

- Simplify $ \frac{6}{4} = \frac{3}{2} $
- LCD of 2 and 12 is 12
- $ \frac{3}{2} = \frac{18}{12} $
- $ \frac{18}{12} + \frac{5}{12} = \frac{23}{12} $
- Simplify: $ \frac{23}{12} = 1\frac{11}{12} $

✔ Answer: $ 1\frac{11}{12} $

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🔹 Problem 11: $ \frac{4}{6} + \frac{2}{9} $

- Simplify $ \frac{4}{6} = \frac{2}{3} $
- LCD of 3 and 9 is 9
- $ \frac{2}{3} = \frac{6}{9} $
- $ \frac{6}{9} + \frac{2}{9} = \frac{8}{9} $

✔ Answer: $ \frac{8}{9} $

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🔹 Problem 12: $ \frac{9}{2} + \frac{2}{8} $

- Simplify $ \frac{2}{8} = \frac{1}{4} $
- LCD of 2 and 4 is 4
- $ \frac{9}{2} = \frac{18}{4} $
- $ \frac{18}{4} + \frac{1}{4} = \frac{19}{4} $
- Simplify: $ \frac{19}{4} = 4\frac{3}{4} $

✔ Answer: $ 4\frac{3}{4} $

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✔ Final Answers:

| # | Answer |
|---|------------------|
| 1 | $ \frac{3}{4} $ |
| 2 | $ 1\frac{1}{8} $ |
| 3 | $ 1\frac{1}{3} $ |
| 4 | $ \frac{5}{6} $ |
| 5 | $ 3\frac{7}{15} $ |
| 6 | $ \frac{9}{10} $ |
| 7 | $ 1\frac{1}{6} $ |
| 8 | $ 1\frac{7}{24} $ |
| 9 | $ \frac{2}{3} $ |
|10 | $ 1\frac{11}{12} $ |
|11 | $ \frac{8}{9} $ |
|12 | $ 4\frac{3}{4} $ |

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