Adding and Subtracting Fractions: Practice

Since $4$ is a multiple of $2$, the LCD is $4$. Only $\frac{1}{2}$ needs rewriting.

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

Now add over the common denominator.

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

Adding the tops and bottoms separately gives $\frac{2}{6}$, which is wrong.

Since $6$ is a multiple of $3$, the LCD is $6$. Rewrite $\frac{2}{3}$ over $6$.

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

Now add over the common denominator.

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

Since $5$ and $6$ share no factor above $1$, $\frac{5}{6}$ is in lowest terms.

Since $6$ is a multiple of $3$, the LCD is $6$. Rewrite $\frac{1}{3}$ over $6$.

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

Now subtract and simplify.

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

Subtracting the tops and bottoms separately gives $\frac{4}{3}$, which is wrong.

Write the whole number $1$ as a fraction over $5$ so the pieces match: $1 = \frac{5}{5}$.

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

Since $3$ and $5$ share no factor above $1$, $\frac{3}{5}$ is in lowest terms.

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

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

Now add over the common denominator.

$\frac{8}{12} + \frac{3}{12} = \frac{11}{12}$

Adding the tops and bottoms separately gives $\frac{3}{7}$, which is wrong.

Since $6$ is a multiple of $2$, the LCD is $6$. Rewrite $\frac{1}{2}$ over $6$.

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

Now add and simplify.

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

Adding the tops and bottoms separately gives $\frac{2}{8}$, which is wrong.

Since $8$ is a multiple of $4$, the LCD is $8$. Rewrite $\frac{3}{4}$ over $8$.

$\frac{3}{4} = \frac{6}{8}$

Now add over the common denominator.

$\frac{6}{8} + \frac{1}{8} = \frac{7}{8}$

Since $7$ and $8$ share no factor above $1$, $\frac{7}{8}$ is in lowest terms.

Since $6$ is a multiple of $2$, the LCD is $6$. Rewrite $\frac{1}{2}$ over $6$.

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

Now subtract and simplify.

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

Subtracting the tops and bottoms separately gives $\frac{4}{4}$, which is wrong.

The least common denominator of $2$ and $5$ is $10$. Rewrite each fraction over $10$.

$\frac{1}{2} = \frac{5}{10}, \qquad \frac{2}{5} = \frac{4}{10}$

Now add over the common denominator.

$\frac{5}{10} + \frac{4}{10} = \frac{9}{10}$

Adding the tops and bottoms separately gives $\frac{3}{7}$, which is wrong.

Since $8$ is a multiple of $2$, the LCD is $8$. Rewrite $\frac{1}{2}$ over $8$.

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

Now subtract over the common denominator.

$\frac{7}{8} - \frac{4}{8} = \frac{3}{8}$

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

Her total is $\frac{1}{4} + \frac{3}{8}$. Since $8$ is a multiple of $4$, the LCD is $8$. Rewrite $\frac{1}{4}$ over $8$.

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

Now add over the common denominator.

$\frac{2}{8} + \frac{3}{8} = \frac{5}{8}$

So Maria ate $\frac{5}{8}$ of the pizza.

What is left is $\frac{5}{6} - \frac{1}{4}$. The least common denominator of $6$ and $4$ is $12$. Rewrite each fraction over $12$.

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

Now subtract over the common denominator.

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

So $\frac{7}{12}$ of the tank is left.