Adding and Subtracting Fractions: Practice

The missing piece is $\frac{3}{4} - \frac{1}{4}$.

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

Check: $\frac{1}{4} + \frac{1}{2} = \frac{1}{4} + \frac{2}{4} = \frac{3}{4}$, so the missing number is $\frac{1}{2}$.

First add inside the parentheses over the denominator $6$.

$\frac{1}{2} + \frac{1}{6} = \frac{3}{6} + \frac{1}{6} = \frac{4}{6}$

Then subtract that from $\frac{5}{6}$.

$\frac{5}{6} - \frac{4}{6} = \frac{1}{6}$

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

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

Now subtract over the common denominator.

$\frac{9}{12} - \frac{8}{12} = \frac{1}{12}$

Subtracting the tops and bottoms separately gives $1$, which is wrong.

The least common denominator of $2$, $4$, and $8$ is $8$. Rewrite each fraction over $8$.

$\frac{1}{2} = \frac{4}{8}, \qquad \frac{1}{4} = \frac{2}{8}, \qquad \frac{1}{8} = \frac{1}{8}$

Now add all three over the common denominator.

$\frac{4}{8} + \frac{2}{8} + \frac{1}{8} = \frac{7}{8}$

Adding all the tops and bottoms separately gives $\frac{3}{14}$, which is wrong.

What remains is $\frac{7}{8} - \frac{1}{4}$. Since $8$ is a multiple of $4$, the LCD is $8$. Rewrite $\frac{1}{4}$ over $8$.

$\frac{1}{4} = \frac{2}{8}$

Now subtract over the common denominator.

$\frac{7}{8} - \frac{2}{8} = \frac{5}{8}$

So $\frac{5}{8}$ m remains.

The missing piece is $\frac{5}{6} - \frac{1}{2}$.

$\frac{5}{6} - \frac{1}{2} = \frac{5}{6} - \frac{3}{6} = \frac{2}{6} = \frac{1}{3}$

Check: $\frac{5}{6} - \frac{1}{3} = \frac{5}{6} - \frac{2}{6} = \frac{3}{6} = \frac{1}{2}$, so the missing number is $\frac{1}{3}$.

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

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

Now add over the common denominator.

$\frac{10}{12} + \frac{9}{12} = \frac{19}{12}$

This improper fraction is the final answer. Adding the tops and bottoms separately gives $\frac{8}{10}$, which is wrong.

The total is $\frac{1}{4} + \frac{1}{3} + \frac{1}{6}$. The LCD of $4$, $3$, and $6$ is $12$. Rewrite each over $12$.

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

Now add and simplify.

$\frac{3}{12} + \frac{4}{12} + \frac{2}{12} = \frac{9}{12} = \frac{3}{4}$

Since $12$ is a multiple of $4$, the LCD is $12$. Rewrite $\frac{1}{4}$ over $12$.

$\frac{1}{4} = \frac{3}{12}$

Now subtract and simplify.

$\frac{7}{12} - \frac{3}{12} = \frac{4}{12} = \frac{1}{3}$

The greatest common factor of $4$ and $12$ is $4$, so $\frac{4}{12} = \frac{1}{3}$.

You need $\frac{3}{4} - \frac{1}{3}$. The least common denominator of $4$ and $3$ is $12$. Rewrite each over $12$.

$\frac{3}{4} = \frac{9}{12}, \qquad \frac{1}{3} = \frac{4}{12}$

Now subtract over the common denominator.

$\frac{9}{12} - \frac{4}{12} = \frac{5}{12}$

So you need $\frac{5}{12}$ cup more.

Write the whole number $1$ as a fraction over $8$ so the pieces match: $1 = \frac{8}{8}$.

$1 - \frac{5}{8} = \frac{8}{8} - \frac{5}{8} = \frac{3}{8}$

Since $3$ and $8$ share no factor above $1$, $\frac{3}{8}$ is in lowest terms.

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

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

Then combine left to right and simplify.

$\frac{8}{12} - \frac{3}{12} + \frac{1}{12} = \frac{6}{12} = \frac{1}{2}$