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

Let’s solve each problem one by one. We’ll work step by step, find common denominators when needed, and simplify to lowest terms.

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

First, simplify inside the parentheses:
Convert mixed numbers to improper fractions:
$4\frac{1}{3} = \frac{13}{3}$, $\frac{3}{2}$ stays as is.

Now subtract:
$\frac{13}{3} - \frac{3}{2}$ → LCD of 3 and 2 is 6
= $\frac{26}{6} - \frac{9}{6} = \frac{17}{6}$

Now do: $2\frac{5}{6} - \frac{17}{6}$
Convert $2\frac{5}{6} = \frac{17}{6}$

So: $\frac{17}{6} - \frac{17}{6} = 0$

✔ Answer: 0

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

Find LCD of 2, 8, 12 → LCD = 24

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

Add first two: $\frac{12}{24} + \frac{39}{24} = \frac{51}{24}$
Subtract third: $\frac{51}{24} - \frac{22}{24} = \frac{29}{24}$

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

✔ Answer: $1\frac{5}{24}$

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

Convert mixed number: $3\frac{4}{5} = \frac{19}{5}$

LCD of 10, 6, 5 → LCD = 30

Convert:
$\frac{3}{10} = \frac{9}{30}$
$\frac{1}{6} = \frac{5}{30}$
$\frac{19}{5} = \frac{114}{30}$

Now compute:
$\frac{9}{30} - \frac{5}{30} + \frac{114}{30} = \frac{4}{30} + \frac{114}{30} = \frac{118}{30}$

Simplify: divide numerator and denominator by 2 → $\frac{59}{15} = 3\frac{14}{15}$

✔ Answer: $3\frac{14}{15}$

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

Notice: $+\frac{2}{7} - \frac{2}{7} = 0$, so we’re left with $\frac{3}{4}$

✔ Answer: $\frac{3}{4}$

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

Combine the last two terms: $\frac{17}{2} - \frac{3}{2} = \frac{14}{2} = 7$

Now add: $1\frac{1}{5} + 7 = 8\frac{1}{5}$

✔ Answer: $8\frac{1}{5}$

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

Convert all to sixths:
$\frac{17}{6}$ stays
$\frac{5}{3} = \frac{10}{6}$
$3\frac{1}{2} = \frac{7}{2} = \frac{21}{6}$

Now: $\frac{17}{6} + \frac{10}{6} = \frac{27}{6}$
Then: $\frac{27}{6} - \frac{21}{6} = \frac{6}{6} = 1$

✔ Answer: 1

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

Convert mixed number: $1\frac{7}{9} = \frac{16}{9}$

LCD of 2, 9, 3 → LCD = 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:
$1\frac{11}{12} - \left(1\frac{3}{4} - \frac{1}{8}\right)$

Inside parentheses: convert to eighths or twelfths? Let’s use 24 for safety.

$1\frac{3}{4} = \frac{7}{4} = \frac{42}{24}$
$\frac{1}{8} = \frac{3}{24}$
So: $\frac{42}{24} - \frac{3}{24} = \frac{39}{24}$

Now: $1\frac{11}{12} = \frac{23}{12} = \frac{46}{24}$

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

✔ Answer: $\frac{7}{24}$

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Problem 9:
$\frac{11}{2} - \left(\frac{2}{7} + \frac{3}{2}\right)$

Inside parentheses: $\frac{2}{7} + \frac{3}{2}$ → LCD = 14
= $\frac{4}{14} + \frac{21}{14} = \frac{25}{14}$

Now: $\frac{11}{2} = \frac{77}{14}$
Subtract: $\frac{77}{14} - \frac{25}{14} = \frac{52}{14} = \frac{26}{7} = 3\frac{5}{7}$

✔ Answer: $3\frac{5}{7}$

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Problem 10:
$3\frac{1}{3} + 1\frac{3}{4} - 1\frac{2}{3}$

Group like terms: $(3\frac{1}{3} - 1\frac{2}{3}) + 1\frac{3}{4}$

$3\frac{1}{3} - 1\frac{2}{3} = (3-1) + (\frac{1}{3}-\frac{2}{3}) = 2 - \frac{1}{3} = 1\frac{2}{3}$

Now add $1\frac{2}{3} + 1\frac{3}{4}$

Convert to twelfths:
$1\frac{2}{3} = \frac{5}{3} = \frac{20}{12}$
$1\frac{3}{4} = \frac{7}{4} = \frac{21}{12}$
Sum: $\frac{41}{12} = 3\frac{5}{12}$

✔ Answer: $3\frac{5}{12}$

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

Inside parentheses: convert to twelfths
$1\frac{11}{12} = \frac{23}{12}$
$\frac{5}{4} = \frac{15}{12}$
So: $\frac{23}{12} - \frac{15}{12} = \frac{8}{12} = \frac{2}{3}$

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

✔ Answer: $\frac{2}{3}$

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Problem 12:
$2\frac{1}{3} - \frac{2}{3} + 1\frac{4}{5}$

First: $2\frac{1}{3} - \frac{2}{3} = 1\frac{2}{3}$ (since $2\frac{1}{3} = \frac{7}{3}, \frac{7}{3} - \frac{2}{3} = \frac{5}{3} = 1\frac{2}{3}$)

Now add $1\frac{2}{3} + 1\frac{4}{5}$

Convert to fifteenths:
$1\frac{2}{3} = \frac{5}{3} = \frac{25}{15}$
$1\frac{4}{5} = \frac{9}{5} = \frac{27}{15}$
Sum: $\frac{52}{15} = 3\frac{7}{15}$

✔ Answer: $3\frac{7}{15}$

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Final Answer:
1. 0
2. $1\frac{5}{24}$
3. $3\frac{14}{15}$
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. $3\frac{7}{15}$