Adding and Subtracting Fractions with Three Terms (A) - Free Printable

Let's solve each of the 12 problems step by step. We'll convert mixed numbers to improper fractions, find common denominators, perform the operations, and simplify to lowest terms.

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Problem 1:

$ 2\frac{5}{6} - \left(4\frac{1}{3} - \frac{3}{2}\right) $

Step 1: Convert all mixed numbers to improper fractions:
- $ 2\frac{5}{6} = \frac{17}{6} $
- $ 4\frac{1}{3} = \frac{13}{3} $
- $ \frac{3}{2} $ stays as is

Now compute inside the parentheses:
$$
\frac{13}{3} - \frac{3}{2}
$$
LCM of 3 and 2 is 6:
$$
= \frac{26}{6} - \frac{9}{6} = \frac{17}{6}
$$

Now subtract:
$$
\frac{17}{6} - \frac{17}{6} = 0
$$

✔ Answer: 0

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Problem 2:

$ \frac{1}{2} + \frac{13}{8} - \frac{11}{12} $

Find LCM of 2, 8, 12 → LCM = 24

Convert:
- $ \frac{1}{2} = \frac{12}{24} $
- $ \frac{13}{8} = \frac{39}{24} $
- $ \frac{11}{12} = \frac{22}{24} $

Now:
$$
\frac{12}{24} + \frac{39}{24} - \frac{22}{24} = \frac{29}{24}
$$

Simplify: $ \frac{29}{24} = 1\frac{5}{24} $

✔ Answer: $ 1\frac{5}{24} $

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Problem 3:

$ \frac{3}{10} - \frac{1}{6} + 2\frac{4}{5} $

Convert $ 2\frac{4}{5} = \frac{14}{5} $

Now: $ \frac{3}{10} - \frac{1}{6} + \frac{14}{5} $

LCM of 10, 6, 5 = 30

Convert:
- $ \frac{3}{10} = \frac{9}{30} $
- $ \frac{1}{6} = \frac{5}{30} $
- $ \frac{14}{5} = \frac{84}{30} $

Now:
$$
\frac{9}{30} - \frac{5}{30} + \frac{84}{30} = \frac{88}{30}
$$

Simplify: $ \frac{88}{30} = \frac{44}{15} = 2\frac{14}{15} $

✔ Answer: $ 2\frac{14}{15} $

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Problem 4:

$ \frac{3}{4} + \frac{2}{7} - \frac{2}{7} $

Note: $ \frac{2}{7} - \frac{2}{7} = 0 $

So only $ \frac{3}{4} $ remains.

✔ Answer: $ \frac{3}{4} $

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Problem 5:

$ 1\frac{1}{5} + \frac{17}{2} - \frac{3}{2} $

Convert $ 1\frac{1}{5} = \frac{6}{5} $

Now:
$$
\frac{6}{5} + \frac{17}{2} - \frac{3}{2}
$$

First: $ \frac{17}{2} - \frac{3}{2} = \frac{14}{2} = 7 $

Now: $ \frac{6}{5} + 7 = \frac{6}{5} + \frac{35}{5} = \frac{41}{5} = 8\frac{1}{5} $

✔ Answer: $ 8\frac{1}{5} $

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Problem 6:

$ \frac{17}{6} + \frac{5}{3} - 3\frac{1}{2} $

Convert $ 3\frac{1}{2} = \frac{7}{2} $

Now: $ \frac{17}{6} + \frac{5}{3} - \frac{7}{2} $

LCM of 6, 3, 2 = 6

Convert:
- $ \frac{17}{6} $
- $ \frac{5}{3} = \frac{10}{6} $
- $ \frac{7}{2} = \frac{21}{6} $

Now:
$$
\frac{17}{6} + \frac{10}{6} - \frac{21}{6} = \frac{6}{6} = 1
$$

✔ Answer: 1

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Problem 7:

$ \frac{5}{2} + 1\frac{7}{9} + \frac{1}{3} $

Convert $ 1\frac{7}{9} = \frac{16}{9} $

Now: $ \frac{5}{2} + \frac{16}{9} + \frac{1}{3} $

LCM of 2, 9, 3 = 18

Convert:
- $ \frac{5}{2} = \frac{45}{18} $
- $ \frac{16}{9} = \frac{32}{18} $
- $ \frac{1}{3} = \frac{6}{18} $

Add:
$$
\frac{45 + 32 + 6}{18} = \frac{83}{18} = 4\frac{11}{18}
$$

✔ Answer: $ 4\frac{11}{18} $

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Problem 8:

$ 1\frac{11}{12} - \left(1\frac{3}{4} - \frac{1}{8}\right) $

Convert:
- $ 1\frac{11}{12} = \frac{23}{12} $
- $ 1\frac{3}{4} = \frac{7}{4} $
- $ \frac{1}{8} $ stays

Inside parentheses:
$$
\frac{7}{4} - \frac{1}{8} = \frac{14}{8} - \frac{1}{8} = \frac{13}{8}
$$

Now:
$$
\frac{23}{12} - \frac{13}{8}
$$

LCM of 12 and 8 = 24

Convert:
- $ \frac{23}{12} = \frac{46}{24} $
- $ \frac{13}{8} = \frac{39}{24} $

Subtract:
$$
\frac{46}{24} - \frac{39}{24} = \frac{7}{24}
$$

✔ Answer: $ \frac{7}{24} $

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Problem 9:

$ \frac{11}{2} - \left(\frac{2}{7} + \frac{3}{2}\right) $

First, add inside parentheses:
$$
\frac{2}{7} + \frac{3}{2} = \text{LCM of 7 and 2 is 14}
$$
$$
= \frac{4}{14} + \frac{21}{14} = \frac{25}{14}
$$

Now:
$$
\frac{11}{2} - \frac{25}{14}
$$

LCM of 2 and 14 = 14

$ \frac{11}{2} = \frac{77}{14} $

$$
\frac{77}{14} - \frac{25}{14} = \frac{52}{14} = \frac{26}{7} = 3\frac{5}{7}
$$

✔ Answer: $ 3\frac{5}{7} $

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Problem 10:

$ 3\frac{1}{3} + 1\frac{3}{4} - 1\frac{2}{3} $

Convert:
- $ 3\frac{1}{3} = \frac{10}{3} $
- $ 1\frac{3}{4} = \frac{7}{4} $
- $ 1\frac{2}{3} = \frac{5}{3} $

Now:
$$
\frac{10}{3} + \frac{7}{4} - \frac{5}{3}
$$

Group like terms:
$$
\left(\frac{10}{3} - \frac{5}{3}\right) + \frac{7}{4} = \frac{5}{3} + \frac{7}{4}
$$

LCM of 3 and 4 = 12

Convert:
- $ \frac{5}{3} = \frac{20}{12} $
- $ \frac{7}{4} = \frac{21}{12} $

Add:
$$
\frac{20}{12} + \frac{21}{12} = \frac{41}{12} = 3\frac{5}{12}
$$

✔ Answer: $ 3\frac{5}{12} $

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Problem 11:

$ \frac{4}{3} - \left(1\frac{11}{12} - \frac{5}{4}\right) $

Convert:
- $ 1\frac{11}{12} = \frac{23}{12} $
- $ \frac{5}{4} $ stays

Inside parentheses:
$$
\frac{23}{12} - \frac{5}{4} = \text{LCM of 12 and 4 is 12}
$$
$$
= \frac{23}{12} - \frac{15}{12} = \frac{8}{12} = \frac{2}{3}
$$

Now:
$$
\frac{4}{3} - \frac{2}{3} = \frac{2}{3}
$$

✔ Answer: $ \frac{2}{3} $

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Problem 12:

$ 2\frac{1}{3} - \frac{2}{3} + 1\frac{4}{5} $

Convert:
- $ 2\frac{1}{3} = \frac{7}{3} $
- $ \frac{2}{3} $ stays
- $ 1\frac{4}{5} = \frac{9}{5} $

Now:
$$
\frac{7}{3} - \frac{2}{3} = \frac{5}{3}
$$

Then:
$$
\frac{5}{3} + \frac{9}{5}
$$

LCM of 3 and 5 = 15

Convert:
- $ \frac{5}{3} = \frac{25}{15} $
- $ \frac{9}{5} = \frac{27}{15} $

Add:
$$
\frac{25 + 27}{15} = \frac{52}{15} = 3\frac{7}{15}
$$

✔ Answer: $ 3\frac{7}{15} $

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✔ Final Answers:

1. $ 0 $
2. $ 1\frac{5}{24} $
3. $ 2\frac{14}{15} $
4. $ \frac{3}{4} $
5. $ 8\frac{1}{5} $
6. $ 1 $
7. $ 4\frac{11}{18} $
8. $ \frac{7}{24} $
9. $ 3\frac{5}{7} $
10. $ 3\frac{5}{12} $
11. $ \frac{2}{3} $
12. $ 3\frac{7}{15} $

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