Adding and Subtracting Fractions: Additional Practice (Set 1)

The least common denominator of $8$ and $12$ is $24$. Rewrite each fraction over $24$.

$\frac{3}{8} = \frac{9}{24}, \qquad \frac{5}{12} = \frac{10}{24}$

Now add over the common denominator.

$\frac{9}{24} + \frac{10}{24} = \frac{19}{24}$

Since $19$ and $24$ share no factor above $1$, $\frac{19}{24}$ is in lowest terms.

The least common denominator of $3$, $9$, and $6$ is $18$. Rewrite each fraction over $18$.

$\frac{2}{3} = \frac{12}{18}, \qquad \frac{4}{9} = \frac{8}{18}, \qquad \frac{1}{6} = \frac{3}{18}$

Now add all three over the common denominator.

$\frac{12}{18} + \frac{8}{18} + \frac{3}{18} = \frac{23}{18}$

Since $23$ and $18$ share no factor above $1$, this improper fraction is the final answer.

Write the whole number $2$ as a fraction over $4$: $2 = \frac{8}{4}$.

$2 + \frac{3}{4} = \frac{8}{4} + \frac{3}{4} = \frac{11}{4}$

This improper fraction is the final answer.

The missing piece is $\frac{5}{6} - \frac{1}{4}$. The LCD of $6$ and $4$ is $12$.

$\frac{5}{6} = \frac{10}{12}, \qquad \frac{1}{4} = \frac{3}{12}$

Subtract over the common denominator.

$\frac{10}{12} - \frac{3}{12} = \frac{7}{12}$

Check: $\frac{5}{6} - \frac{7}{12} = \frac{10}{12} - \frac{7}{12} = \frac{3}{12} = \frac{1}{4}$, so the missing number is $\frac{7}{12}$.

The least common denominator of $6$ and $8$ is $24$. Rewrite each fraction over $24$.

$\frac{5}{6} = \frac{20}{24}, \qquad \frac{7}{8} = \frac{21}{24}$

Now add over the common denominator.

$\frac{20}{24} + \frac{21}{24} = \frac{41}{24}$

This improper fraction is the final answer.

The least common denominator of $4$, $3$, and $6$ is $12$. Rewrite each fraction over $12$.

$\frac{3}{4} = \frac{9}{12}, \qquad \frac{2}{3} = \frac{8}{12}, \qquad \frac{5}{6} = \frac{10}{12}$

Then combine left to right over the common denominator.

$\frac{9}{12} + \frac{8}{12} - \frac{10}{12} = \frac{7}{12}$

The least common denominator of $10$ and $6$ is $30$. Rewrite each fraction over $30$.

$\frac{3}{10} = \frac{9}{30}, \qquad \frac{5}{6} = \frac{25}{30}$

Now add and simplify.

$\frac{9}{30} + \frac{25}{30} = \frac{34}{30} = \frac{17}{15}$

The greatest common factor of $34$ and $30$ is $2$, so $\frac{34}{30} = \frac{17}{15}$.

Rewrite both over the least common denominator $18$.

$\frac{5}{6} = \frac{15}{18}, \qquad \frac{7}{9} = \frac{14}{18}$

With the same denominator, the larger numerator wins, so $\frac{5}{6}$ is larger. The gap is the difference.

$\frac{15}{18} - \frac{14}{18} = \frac{1}{18}$

So $\frac{5}{6}$ is larger by $\frac{1}{18}$.

The least common denominator of $12$, $8$, and $4$ is $24$. Rewrite each fraction over $24$.

$\frac{11}{12} = \frac{22}{24}, \qquad \frac{3}{8} = \frac{9}{24}, \qquad \frac{1}{4} = \frac{6}{24}$

Then subtract left to right over the common denominator.

$\frac{22}{24} - \frac{9}{24} - \frac{6}{24} = \frac{7}{24}$

The runner has covered $\frac{2}{5} + \frac{1}{4}$. The LCD of $5$ and $4$ is $20$.

$\frac{2}{5} = \frac{8}{20}, \qquad \frac{1}{4} = \frac{5}{20}, \qquad \frac{8}{20} + \frac{5}{20} = \frac{13}{20}$

What is left is $1$ minus that, with $1 = \frac{20}{20}$.

$\frac{20}{20} - \frac{13}{20} = \frac{7}{20}$

So $\frac{7}{20}$ of the race is left.

The least common denominator of $5$, $3$, and $15$ is $15$. Rewrite each fraction over $15$.

$\frac{4}{5} = \frac{12}{15}, \qquad \frac{1}{3} = \frac{5}{15}, \qquad \frac{1}{15} = \frac{1}{15}$

Then combine left to right over the common denominator.

$\frac{12}{15} - \frac{5}{15} + \frac{1}{15} = \frac{8}{15}$

Since $8$ and $15$ share no factor above $1$, $\frac{8}{15}$ is in lowest terms.

The least common denominator of $6$, $10$, and $15$ is $30$. Rewrite each fraction over $30$.

$\frac{5}{6} = \frac{25}{30}, \qquad \frac{3}{10} = \frac{9}{30}, \qquad \frac{1}{15} = \frac{2}{30}$

Now add all three and simplify.

$\frac{25}{30} + \frac{9}{30} + \frac{2}{30} = \frac{36}{30} = \frac{6}{5}$

The greatest common factor of $36$ and $30$ is $6$, so $\frac{36}{30} = \frac{6}{5}$.