2.2 Add and Subtract Fractions

Pooja Gupta

2.2 Add and Subtract Fractions

Learning Objectives

By the end of this section, you will be able to:

Add or subtract fractions with a common denominator
Add or subtract fractions with different denominators
Use the order of operations to simplify complex fractions
Evaluate variable expressions with fractions

Add or Subtract Fractions with a Common Denominator

When we multiplied fractions, we just multiplied the numerators and multiplied the denominators right straight across. To add or subtract fractions, they must have a common denominator.

Fraction Addition and Subtraction

If $a,b,\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}c$ are numbers where $c\ne 0$ , then

$\dfrac{a}{c}+\dfrac{b}{c}=\dfrac{a+b}{c}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\dfrac{a}{c}-\phantom{\rule{0.2em}{0ex}}\dfrac{b}{c}=\dfrac{a-b}{c}$

To add or subtract fractions, add or subtract the numerators and place the result over the common denominator.

EXAMPLE 1

Find the sum: $\dfrac{x}{3}+\dfrac{2}{3}$ .

Solution

	$\dfrac{x}{3}+\dfrac{2}{3}$
Add the numerators and place the sum over the common denominator.	$\dfrac{x+2}{3}$

TRY IT 1.1

Find the sum: $\dfrac{x}{4}+\dfrac{3}{4}$ .

Answer

$\frac{x+3}{4}$

TRY IT 1.2

Find the sum: $\dfrac{y}{8}+\dfrac{5}{8}$ .

Answer

$\frac{y+5}{8}$

EXAMPLE 2

Find the difference: $-\phantom{\rule{0.2em}{0ex}}\dfrac{23}{24}-\phantom{\rule{0.2em}{0ex}}\dfrac{13}{24}$ .

Solution

	$-\phantom{\rule{0.2em}{0ex}}\dfrac{23}{24}-\phantom{\rule{0.2em}{0ex}}\dfrac{13}{24}$
Subtract the numerators and place the difference over the common denominator.	$\dfrac{-23-13}{24}$
Simplify.	$\dfrac{-36}{24}$
Simplify. Remember, $-\phantom{\rule{0.2em}{0ex}}\frac{a}{b}=\frac{-a}{b}$ .	$-\phantom{\rule{0.2em}{0ex}}\dfrac{3}{2}$

TRY IT 2.1

Find the difference: $-\phantom{\rule{0.2em}{0ex}}\dfrac{19}{28}-\phantom{\rule{0.2em}{0ex}}\dfrac{7}{28}$ .

Answer

$-\phantom{\rule{0.2em}{0ex}}\frac{26}{28}=-\frac{13}{14}$

TRY IT 2.2

Find the difference: $-\phantom{\rule{0.2em}{0ex}}\dfrac{27}{32}-\phantom{\rule{0.2em}{0ex}}\dfrac{1}{32}$ .

Answer

$-\phantom{\rule{0.2em}{0ex}}\frac{7}{8}$

EXAMPLE 3

Simplify: $-\phantom{\rule{0.2em}{0ex}}\dfrac{10}{x}-\phantom{\rule{0.2em}{0ex}}\dfrac{4}{x}$ .

Solution

	$-\phantom{\rule{0.2em}{0ex}}\dfrac{10}{x}-\phantom{\rule{0.2em}{0ex}}\dfrac{4}{x}$
Subtract the numerators and place the difference over the common denominator.	$\dfrac{-14}{x}$
Rewrite with the sign in front of the fraction.	$-\phantom{\rule{0.2em}{0ex}}\dfrac{14}{x}$

TRY IT 3.1

Find the difference: $-\phantom{\rule{0.2em}{0ex}}\dfrac{9}{x}-\phantom{\rule{0.2em}{0ex}}\dfrac{7}{x}$ .

Answer

$-\phantom{\rule{0.2em}{0ex}}\frac{16}{x}$

TRY IT 3.2

Find the difference: $-\phantom{\rule{0.2em}{0ex}}\dfrac{17}{a}-\phantom{\rule{0.2em}{0ex}}\dfrac{5}{a}$ .

Answer

$-\phantom{\rule{0.2em}{0ex}}\frac{22}{a}$

Now we will do an example that has both addition and subtraction.

EXAMPLE 4

Simplify: $\dfrac{3}{8}+\left(-\phantom{\rule{0.2em}{0ex}}\dfrac{5}{8}\right)-\phantom{\rule{0.2em}{0ex}}\dfrac{1}{8}$ .

Solution

Add and subtract fractions—do they have a common denominator? Yes.	$\dfrac{3}{8}+\left(-\phantom{\rule{0.2em}{0ex}}\dfrac{5}{8}\right)-\phantom{\rule{0.2em}{0ex}}\dfrac{1}{8}$
Add and subtract the numerators and place the difference over the common denominator.	$\dfrac{3+\left(-5\right)-1}{8}$
Simplify left to right.	$\dfrac{-2-1}{8}$
Simplify.	$-\phantom{\rule{0.2em}{0ex}}\dfrac{3}{8}$

TRY IT 4.1

Simplify: $\dfrac{2}{9}+\left(-\phantom{\rule{0.2em}{0ex}}\dfrac{4}{9}\right)-\phantom{\rule{0.2em}{0ex}}\dfrac{7}{9}$ .

Answer

$-\phantom{\rule{0.2em}{0ex}}{1}$

TRY IT 4.2

Simplify: $\dfrac{5}{9}+\left(-\phantom{\rule{0.2em}{0ex}}\dfrac{4}{9}\right)-\phantom{\rule{0.2em}{0ex}}\dfrac{7}{9}$ .

Answer

$-\phantom{\rule{0.2em}{0ex}}\frac{2}{3}$

Add or Subtract Fractions with Different Denominators

As we have seen, to add or subtract fractions, their denominators must be the same. The least common denominator (LCD) of two fractions is the smallest number that can be used as a common denominator of the fractions. The LCD of the two fractions is the least common multiple (LCM) of their denominators.

Least Common Denominator

The least common denominator (LCD) of two fractions is the least common multiple (LCM) of their denominators.

After we find the least common denominator of two fractions, we convert the fractions to equivalent fractions with the LCD. Putting these steps together allows us to add and subtract fractions because their denominators will be the same!

EXAMPLE 5

Add: $\dfrac{7}{12}+\dfrac{5}{18}$ .

Solution

Step 1. Do they have a common denominator?
No-rewrite each fraction with the LCD (least common denominator).

Find the LCD of 12, 18.

Change into equivalent fractions with the LCD, 36.
Do not simplify the equivalent fractions! If you do, you’ll get back to the original fractions and lose the common denominator!

$\begin{aligned} 12 & =2 \cdot 2 \cdot 3 \\ 18 & =2 \cdot \quad 3 \cdot 3 \\ \hline \mathrm{LCD} & =2 \cdot 2 \cdot 3 \cdot 3 \\ \mathrm{LCD} & =36 \\ \\ \frac{7}{12} & +\frac{5}{18} \\ \frac{7 \cdot 3}{12 \cdot 3} & +\frac{5 \cdot 2}{18 \cdot 2} \\ \frac{21}{36} & +\frac{10}{36} \end{aligned}$

Step 2. Add or subtract the fractions.

Add.

$\dfrac{31}{36}$

Step 3. Simplify, if possible.

Because 31 is a prime number, it has no factors in common with 36. The answer is simplified.

TRY IT 5.1

Add: $\dfrac{7}{12}+\dfrac{11}{15}$ .

Answer

$\frac{79}{60}$

TRY IT 5.2

Add: $\dfrac{13}{15}+\dfrac{17}{20}$ .

Answer

$\frac{103}{60}$

HOW TO: Add or Subtract Fractions

Do they have a common denominator?
- Yes—go to step 2.
- No—rewrite each fraction with the LCD (least common denominator). Find the LCD. Change each fraction into an equivalent fraction with the LCD as its denominator.
Add or subtract the fractions.
Simplify, if possible.

When finding the equivalent fractions needed to create the common denominators, there is a quick way to find the number we need to multiply both the numerator and denominator. This method works if we found the LCD by factoring into primes.

Look at the factors of the LCD and then at each column above those factors. The “missing” factors of each denominator are the numbers we need.

In (Example 5), the LCD, 36, has two factors of 2 and two factors of $3$ .

The numerator 12 has two factors of 2 but only one of 3—so it is “missing” one 3—we multiply the numerator and denominator by 3

The numerator 18 is missing one factor of 2—so we multiply the numerator and denominator by 2

We will apply this method as we subtract the fractions in (Example 6).

EXAMPLE 6

Subtract: $\dfrac{7}{15}-\phantom{\rule{0.2em}{0ex}}\dfrac{19}{24}$ .

Solution

Do the fractions have a common denominator? No, so we need to find the LCD.

Find the LCD. $\phantom{\rule{5em}{0ex}}$	$\begin{aligned} & \frac{7}{15}-\frac{19}{24} \\ 15&= \quad \quad \quad \quad 3 \cdot 5 \\ 24&= 2 \cdot 2 \cdot 2 \cdot 3 \\ \hline \text{LCD}&=2 \cdot 2 \cdot 2 \cdot 3 \cdot 5 \\ \text{LCD}&=120 \end{aligned}$
Notice, 15 is “missing” three factors of 2 and 24 is “missing” the 5 from the factors of the LCD. So we multiply 8 in the first fraction and 5 in the second fraction to get the LCD.
Rewrite as equivalent fractions with the LCD.	$\dfrac{7 \cdot {\color{red}8}}{15 \cdot {\color{red}8}}-\dfrac{19 \cdot {\color{red}5}}{24 \cdot {\color{red}5}}$
Simplify.	$\dfrac{56}{120}-\dfrac{95}{120}$
Subtract.	$-\dfrac{39}{120}$
Check to see if the answer can be simplified. Both 39 and 120 have a factor of 3.	$-\dfrac{13\cdot 3}{40\cdot 3}$
Simplify.	$-\dfrac{13}{40}$

Do not simplify the equivalent fractions! If you do, you’ll get back to the original fractions and lose the common denominator!

TRY IT 6.1

Subtract: $\dfrac{13}{24}-\phantom{\rule{0.2em}{0ex}}\dfrac{17}{32}$ .

Answer

$\frac{1}{96}$

TRY IT 6.2

Subtract: $\dfrac{21}{32}-\phantom{\rule{0.2em}{0ex}}\dfrac{9}{28}$ .

Answer

$\frac{75}{224}$

In the next example, one of the fractions has a variable in its numerator. Notice that we do the same steps as when both numerators are numbers.

EXAMPLE 7

Add: $\dfrac{3}{5}+\dfrac{x}{8}$ .

Solution

The fractions have different denominators.

Find the LCD.	$\begin{aligned} & \frac{3}{5}+\frac{x}{8} \\ 5&= \quad \quad \quad 5 \\ 8&= 2 \cdot 2 \cdot 2 \\ \hline \text{LCD}&=2 \cdot 2 \cdot 2 \cdot 5 \\ \text{LCD}&=40 \end{aligned}$
Rewrite as equivalent fractions with the LCD.	$\dfrac{3 \cdot {\color{red}{8}}}{5 \cdot {\color{red}{8}}}+\dfrac{x \cdot {\color{red}{5}}}{8 \cdot {\color{red}{5}}}$
Simplify.	$\dfrac{24}{40}+\dfrac{5 x}{40}$
Add.	$\dfrac{24+5x}{40}$

TRY IT 7.1

Add: $\dfrac{y}{6}+\dfrac{7}{9}$ .

Answer

$\frac{9y+42}{54}$

TRY IT 7.2

Add: $\dfrac{x}{6}+\dfrac{7}{15}$ .

Answer

$\frac{15x+42}{90}$

We now have all four operations for fractions. The table below summarizes fraction operations.

Summary of Fraction Operations
Fraction Operation	Sample Equation	What to Do
Fraction multiplication	$\frac{a}{b} \cdot \frac{c}{d}=\frac{ac}{bd}$	Multiply the numerators and multiply the denominators
Fraction division	$\frac{a}{b} \div \frac{c}{d}=\frac{a}{b}\cdot \frac{d}{c}$	Multiply the first fraction by the reciprocal of the second.
Fraction addition	$\frac{a}{c}+\frac{b}{c}=\frac{a+b}{c}$	Add the numerators and place the sum over the common denominator.
Fraction subtraction	$\frac{a}{c}-\phantom{\rule{0.2em}{0ex}}\frac{b}{c}=\frac{a-b}{c}$	Subtract the numerators and place the difference over the common denominator.

To multiply or divide fractions, an LCD is NOT needed. To add or subtract fractions, an LCD is needed.

EXAMPLE 8

Simplify: a) $\dfrac{5x}{6}-\phantom{\rule{0.2em}{0ex}}\dfrac{3}{10}$ b) $\dfrac{5x}{6}\cdot\dfrac{3}{10}$ .

Solution

First ask, “What is the operation?” Once we identify the operation that will determine whether we need a common denominator. Remember, we need a common denominator to add or subtract, but not to multiply or divide.

a) What is the operation? The operation is subtraction.
Do the fractions have a common denominator? No.	$\dfrac{5x}{6}-\dfrac{3}{10}$
Rewrite each fraction as an equivalent fraction with the LCD.	$\dfrac{5x\cdot5}{6\cdot5}-\dfrac{3\cdot3}{10\cdot3}$ $\dfrac{25x}{30}-\dfrac{9}{30}$
Subtract the numerators and place the difference over the common denominators.	$\dfrac{25x-9}{30}$
Simplify, if possible.	There are no common factors. The fraction is simplified.

b) What is the operation? Multiplication.	$\dfrac{5x}{6}\cdot\dfrac{3}{10}$
To multiply fractions, multiply the numerators and multiply the denominators.	$\dfrac{5x\cdot3}{6\cdot10}$
Rewrite, showing common factors. Remove common factors.	$\dfrac{\cancel{5}x \cdot \cancel{3}}{2\cdot\cancel{3}\cdot2\cdot\cancel{5}}$
Simplify.	$\dfrac{x}{4}$

Notice we needed an LCD to add $\dfrac{5x}{6}-\phantom{\rule{0.2em}{0ex}}\dfrac{3}{10}$ , but not to multiply $\dfrac{5x}{6}\cdot\dfrac{3}{10}$ .

TRY IT 8.1

Simplify. a) $\dfrac{3a}{4}- \dfrac{8}{9}$ b) $\dfrac{3a}{4} \cdot \dfrac{8}{9}$

Answer

a) $\frac{27a-32}{36}$ b) $\frac{2a}{3}$

TRY IT 8.2

Simplify: a) $\dfrac{4k}{5}-\phantom{\rule{0.2em}{0ex}}\dfrac{1}{6}$ b) $\dfrac{4k}{5}\cdot \dfrac{1}{6}$ .

Answer

a) $\frac{24k-5}{30}$ b) $\frac{2k}{15}$

Use the Order of Operations to Simplify Complex Fractions

We have seen that a complex fraction is a fraction in which the numerator or denominator contains a fraction. The fraction bar indicates division. We simplified the complex fraction $\frac{\frac{3}{4}}{\frac{5}{8}}$ by dividing $\frac{3}{4}$ by $\frac{5}{8}$ .

Now we’ll look at complex fractions where the numerator or denominator contains an expression that can be simplified. So we first must completely simplify the numerator and denominator separately using the order of operations. Then we divide the numerator by the denominator.

EXAMPLE 9

Simplify: $\dfrac{{\left(\dfrac{1}{2}\right)}^{2}}{4+{3}^{2}}$ .

Solution

Step 1. Simplify the numerator.	* Remember, $\left(\frac{1}{2}\right)^2$ means $\frac{1}{2} \cdot \frac{1}{2}$	$\begin{aligned} & \frac{\left(\frac{1}{2}\right)^2}{4+3^2} &= \frac{\frac{1}{4}}{4+3^2} \end{aligned}$
Step 2. Simplify the denominator.		$\begin{aligned} & \frac{\frac{1}{4}}{4+9} &= \frac{\frac{1}{4}}{13} \end{aligned}$
Step 3. Divide the numerator by the denominator. Simplify, if possible.	* Remember, $13=\frac{13}{1}$	$\begin{aligned} \frac{1}{4} &\div 13 =\frac{1}{4} \cdot \frac{1}{13} = \frac{1}{52} \end{aligned}$

TRY IT 9.1

Simplify: $\dfrac{{\left(\dfrac{1}{3}\right)}^{2}}{{2}^{3}+2}$ .

Answer

$\frac{1}{90}$

TRY IT 9.2

Simplify: $\dfrac{1+{4}^{2}}{{\left(\dfrac{1}{4}\right)}^{2}}$ .

Answer

$272$

HOW TO: Simplify Complex Fractions

Simplify the numerator.
Simplify the denominator.
Divide the numerator by the denominator. Simplify if possible.

EXAMPLE 10

Simplify: $\dfrac{\dfrac{1}{2}+\dfrac{2}{3}}{\dfrac{3}{4}-\phantom{\rule{0.2em}{0ex}}\dfrac{1}{6}}$ .

Solution

It may help to put parentheses around the numerator and the denominator.

	$\dfrac{\left(\dfrac{1}{2}+\dfrac{2}{3}\right)}{\left(\dfrac{3}{4}-\phantom{\rule{0.2em}{0ex}}\dfrac{1}{6}\right)}$
Simplify the numerator (LCD = 6) and simplify the denominator (LCD = 12).	$\dfrac{\left(\dfrac{3}{6}+\dfrac{4}{6}\right)}{\left(\dfrac{9}{12}-\phantom{\rule{0.2em}{0ex}}\dfrac{2}{12}\right)}$
Simplify.	$\dfrac{\left(\dfrac{7}{6}\right)}{\left(\dfrac{7}{12}\right)}$
Divide the numerator by the denominator.	$\dfrac{7}{6}\div \dfrac{7}{12}$
Simplify.	$\dfrac{7}{6}\cdot\dfrac{12}{7}$
Divide out common factors.	$\dfrac{7\cdot6\cdot2}{6\cdot7}$
Simplify.	$2$

TRY IT 10.1

Simplify: $\dfrac{\dfrac{1}{3}+\dfrac{1}{2}}{\dfrac{3}{4}-\phantom{\rule{0.2em}{0ex}}\dfrac{1}{3}}$ .

Answer

2

TRY IT 10.2

Simplify: $\dfrac{\dfrac{2}{3}-\phantom{\rule{0.2em}{0ex}}\dfrac{1}{2}}{\dfrac{1}{4}+\dfrac{1}{3}}$ .

Answer

$\frac{2}{7}$

Evaluate Variable Expressions with Fractions

We have evaluated expressions before, but now we can evaluate expressions with fractions. Remember, to evaluate an expression, we substitute the value of the variable into the expression and then simplify.

EXAMPLE 11

Evaluate $x+\dfrac{1}{3}$ when a) $x=-\phantom{\rule{0.2em}{0ex}}\dfrac{1}{3}$ b) $x=-\phantom{\rule{0.2em}{0ex}}\dfrac{3}{4}$ .

Solution

To evaluate $x+\dfrac{1}{3}$ when $x=-\phantom{\rule{0.2em}{0ex}}\dfrac{1}{3}$ , substitute $-\phantom{\rule{0.2em}{0ex}}\dfrac{1}{3}$ for $x$ in the expression. $x+\dfrac{1}{3}$

Substitute $\color{red}-\dfrac{1}{3}$ for $x$ . ${\color{red}-\dfrac{1}{3}}+\dfrac{1}{3}$

Simplify. $\phantom{\rule{18em}{0ex}}$ $0$
To evaluate $x+\dfrac{1}{3}$ when $x=-\phantom{\rule{0.2em}{0ex}}\dfrac{3}{4}$ , we substitute $-\phantom{\rule{0.2em}{0ex}}\dfrac{3}{4}$ for x in the expression. $x+\dfrac{1}{3}$

Substitute $\color{red}-\dfrac{3}{4}$ for $x$ . ${\color{red}-\dfrac{3}{4}}+\dfrac{1}{3}$

Rewrite as equivalent fractions with the LCD, 12. $-\dfrac{3 \cdot {\color{red}3}}{4 \cdot {\color{red}3}}+\dfrac{1 \cdot {\color{red}4}}{3 \cdot {\color{red}4}}$

Simplify. $-\dfrac{9}{12}+\dfrac{4}{12}$

Add. $-\phantom{\rule{0.2em}{0ex}}\dfrac{5}{12}$

TRY IT 11.1

Evaluate $x+\dfrac{3}{4}$ when a) $x=-\phantom{\rule{0.2em}{0ex}}\dfrac{7}{4}$ b) $x=-\phantom{\rule{0.2em}{0ex}}\dfrac{5}{4}$ .

Answer

a) $-1$ b) $-\phantom{\rule{0.2em}{0ex}}\frac{1}{2}$

TRY IT 11.2

Evaluate $y+\dfrac{1}{2}$ when a) $y=\dfrac{2}{3}$ b) $y=-\phantom{\rule{0.2em}{0ex}}\dfrac{3}{4}$ .

Answer

a) $\frac{7}{6}$ b) $-\phantom{\rule{0.2em}{0ex}}\frac{1}{4}$

EXAMPLE 12

Evaluate $-\phantom{\rule{0.2em}{0ex}}\dfrac{5}{6}-y$ when $y=-\phantom{\rule{0.2em}{0ex}}\dfrac{2}{3}$ .

Solution

	$-\frac{5}{6}-y$
Substitute ${\color{red}-\dfrac{2}{3}}$ for $y$ .	$-\dfrac{5}{6}-\left({\color{red}-\dfrac{2}{3}}\right)$
Rewrite as equivalent fractions with the LCD, 6.	$-\dfrac{5}{6}-\left({\color{red}-\dfrac{4}{6}}\right)$
Subtract.	$\dfrac{-5-(-4)}{6} = \dfrac{-5+4}{6}$
Simplify.	$-\phantom{\rule{0.2em}{0ex}}\dfrac{1}{6}$

TRY IT 12.1

Evaluate $-\phantom{\rule{0.2em}{0ex}}\dfrac{1}{2}-y$ when $y=-\phantom{\rule{0.2em}{0ex}}\dfrac{1}{4}$ .

Answer

$-\phantom{\rule{0.2em}{0ex}}\frac{1}{4}$

TRY IT 12.2

Evaluate $-\phantom{\rule{0.2em}{0ex}}\dfrac{3}{8}-y$ when $y=-\phantom{\rule{0.2em}{0ex}}\dfrac{5}{2}$ .

Answer

$\phantom{\rule{0.2em}{0ex}}\frac{17}{8}$

EXAMPLE 13

Evaluate $2{x}^{2}y$ when $x=\dfrac{1}{4}$ and $y=-\phantom{\rule{0.2em}{0ex}}\dfrac{2}{3}$ .

Solution

Substitute the values into the expression.

	$2{x}^{2}y$
Substitute $\color{red}\frac{1}{4}$ for $x$ and $\color{blue}-\frac{2}{3}$ for $y$ .	$2\left({\color{red}\dfrac{1}{4}}\right)^2\left({\color{blue}{-\dfrac{2}{3}}}\right)$
Simplify exponents first.	$2\left(\dfrac{1}{16}\right)\left(-\phantom{\rule{0.2em}{0ex}}\dfrac{2}{3}\right)$
Multiply. Divide out the common factors. Notice we write 16 as $2\cdot 2\cdot 4$ to make it easy to remove common factors.	$- \dfrac{\cancel{2}\cdot 1\cdot \cancel{2}}{\cancel{2}\cdot \cancel{2}\cdot 4\cdot 3}$
Simplify.	$-\phantom{\rule{0.2em}{0ex}}\frac{1}{12}$

TRY IT 13.1

Evaluate $3a{b}^{2}$ when $a=-\phantom{\rule{0.2em}{0ex}}\dfrac{2}{3}$ and $b=-\phantom{\rule{0.2em}{0ex}}\dfrac{1}{2}$ .

Answer

$-\phantom{\rule{0.2em}{0ex}}\frac{1}{2}$

TRY IT 13.2

Evaluate $4{c}^{3}d$ when $c=-\phantom{\rule{0.2em}{0ex}}\dfrac{1}{2}$ and $d=-\phantom{\rule{0.2em}{0ex}}\dfrac{4}{3}$ .

Answer

$\frac{2}{3}$

The next example will have only variables, no constants.

EXAMPLE 14

Evaluate $\dfrac{p+q}{r}$ when $p=-4,q=-2,\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}r=8$ .

Solution

To evaluate $\dfrac{p+q}{r}$ when $p=-4,q=-2,\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}r=8$ , we substitute the values into the expression.

	$\dfrac{p+q}{r}$
Substitute -4 for $p$ , -2 for $q$ and 8 for $r$ .	$\definecolor{darkgreen}{RGB}{0,100,0} \dfrac{{\color{red}-4}+{\color{blue}(-2)}}{\color{darkgreen}{{8}}}$
Add in the numerator first.	$\dfrac{-6}{8}$
Simplify.	$-\dfrac{3}{4}$

TRY IT 14.1

Evaluate $\dfrac{a+b}{c}$ when $a=-8,b=-7,\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}c=6$ .

Answer

$-\phantom{\rule{0.2em}{0ex}}\frac{5}{2}$

TRY IT 14.2

Evaluate $\dfrac{x+y}{z}$ when $x=9,y=-18,\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}z=-6$ .

Answer

$\frac{3}{2}$

Key Concepts

Fraction Addition and Subtraction: If $a,b,\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}c$ are numbers where $c\ne 0$ , then
$\frac{a}{c}+\frac{b}{c}=\frac{a+b}{c}$ and $\frac{a}{c}-\phantom{\rule{0.2em}{0ex}}\frac{b}{c}=\frac{a-b}{c}$ .
To add or subtract fractions, add or subtract the numerators and place the result over the common denominator.
Strategy for Adding or Subtracting Fractions
1. Do they have a common denominator?
  Yes—go to step 2.
  No—Rewrite each fraction with the LCD (Least Common Denominator). Find the LCD. Change each fraction into an equivalent fraction with the LCD as its denominator.
2. Add or subtract the fractions.
3. Simplify, if possible. To multiply or divide fractions, an LCD IS NOT needed. To add or subtract fractions, an LCD IS needed.
Simplify Complex Fractions
1. Simplify the numerator.
2. Simplify the denominator.
3. Divide the numerator by the denominator. Simplify if possible.

Glossary

least common denominator: The least common denominator (LCD) of two fractions is the Least common multiple (LCM) of their denominators.

Practice Makes Perfect

Add and Subtract Fractions with a Common Denominator

In the following exercises, add.

1. $\frac{6}{13}+\frac{5}{13}$	2. $\frac{4}{15}+\frac{7}{15}$
3. $\frac{x}{4}+\frac{3}{4}$	4. $\frac{8}{q}+\frac{6}{q}$
5. $-\phantom{\rule{0.2em}{0ex}}\frac{3}{16}+\left(-\phantom{\rule{0.2em}{0ex}}\frac{7}{16}\right)$	6. $-\phantom{\rule{0.2em}{0ex}}\frac{5}{16}+\left(-\phantom{\rule{0.2em}{0ex}}\frac{9}{16}\right)$
7. $-\phantom{\rule{0.2em}{0ex}}\frac{8}{17}+\frac{15}{17}$	8. $-\phantom{\rule{0.2em}{0ex}}\frac{9}{19}+\frac{17}{19}$
9. $\frac{6}{13}+\left(-\phantom{\rule{0.2em}{0ex}}\frac{10}{13}\right)+\left(-\phantom{\rule{0.2em}{0ex}}\frac{12}{13}\right)$	10. $\frac{5}{12}+\left(-\phantom{\rule{0.2em}{0ex}}\frac{7}{12}\right)+\left(-\phantom{\rule{0.2em}{0ex}}\frac{11}{12}\right)$

In the following exercises, subtract.

11. $\frac{11}{15}-\phantom{\rule{0.2em}{0ex}}\frac{7}{15}$	12. $\frac{9}{13}-\phantom{\rule{0.2em}{0ex}}\frac{4}{13}$
13. $\frac{11}{12}-\phantom{\rule{0.2em}{0ex}}\frac{5}{12}$	14. $\frac{7}{12}-\phantom{\rule{0.2em}{0ex}}\frac{5}{12}$
15. $\frac{19}{21}-\phantom{\rule{0.2em}{0ex}}\frac{4}{21}$	16. $\frac{17}{21}-\phantom{\rule{0.2em}{0ex}}\frac{8}{21}$
17. $\frac{5y}{8}-\phantom{\rule{0.2em}{0ex}}\frac{7}{8}$	18. $\frac{11z}{13}-\phantom{\rule{0.2em}{0ex}}\frac{8}{13}$
19. $-\phantom{\rule{0.2em}{0ex}}\frac{23}{u}-\phantom{\rule{0.2em}{0ex}}\frac{15}{u}$	20. $-\phantom{\rule{0.2em}{0ex}}\frac{29}{v}-\phantom{\rule{0.2em}{0ex}}\frac{26}{v}$
21. $-\phantom{\rule{0.2em}{0ex}}\frac{3}{5}-\left(-\phantom{\rule{0.2em}{0ex}}\frac{4}{5}\right)$	22. $-\phantom{\rule{0.2em}{0ex}}\frac{3}{7}-\left(-\phantom{\rule{0.2em}{0ex}}\frac{5}{7}\right)$
23. $-\phantom{\rule{0.2em}{0ex}}\frac{7}{9}-\left(-\phantom{\rule{0.2em}{0ex}}\frac{5}{9}\right)$	24. $-\phantom{\rule{0.2em}{0ex}}\frac{8}{11}-\left(-\phantom{\rule{0.2em}{0ex}}\frac{5}{11}\right)$

Mixed Practice

In the following exercises, simplify.

25. $-\phantom{\rule{0.2em}{0ex}}\frac{5}{18}\cdot \frac{9}{10}$	26. $-\phantom{\rule{0.2em}{0ex}}\frac{3}{14}\cdot \frac{7}{12}$
27. $\frac{n}{5}-\phantom{\rule{0.2em}{0ex}}\frac{4}{5}$	28. $\frac{6}{11}-\phantom{\rule{0.2em}{0ex}}\frac{s}{11}$
29. $-\phantom{\rule{0.2em}{0ex}}\frac{7}{24}+\frac{2}{24}$	30. $-\phantom{\rule{0.2em}{0ex}}\frac{5}{18}+\frac{1}{18}$
31. $\frac{8}{15}\div \frac{12}{5}$	32. $\frac{7}{12}\div\frac{9}{28}$

Add or Subtract Fractions with Different Denominators

In the following exercises, add or subtract.

33. $\frac{1}{2}+\frac{1}{7}$	34. $\frac{1}{3}+\frac{1}{8}$
35. $\frac{1}{3}-\left(-\phantom{\rule{0.2em}{0ex}}\frac{1}{9}\right)$	36. $\frac{1}{4}-\left(-\phantom{\rule{0.2em}{0ex}}\frac{1}{8}\right)$
37. $\frac{7}{12}+\frac{5}{8}$	38. $\frac{5}{12}+\frac{3}{8}$
39. $\frac{7}{12}-\phantom{\rule{0.2em}{0ex}}\frac{9}{16}$	40. $\frac{7}{16}-\phantom{\rule{0.2em}{0ex}}\frac{5}{12}$
41. $\frac{2}{3}-\phantom{\rule{0.2em}{0ex}}\frac{3}{8}$	42. $\frac{5}{6}-\phantom{\rule{0.2em}{0ex}}\frac{3}{4}$
43. $-\phantom{\rule{0.2em}{0ex}}\frac{11}{30}+\frac{27}{40}$	44. $-\phantom{\rule{0.2em}{0ex}}\frac{9}{20}+\frac{17}{30}$
45. $-\phantom{\rule{0.2em}{0ex}}\frac{13}{30}+\frac{25}{42}$	46. $-\phantom{\rule{0.2em}{0ex}}\frac{23}{30}+\frac{5}{48}$
47. $-\phantom{\rule{0.2em}{0ex}}\frac{39}{56}-\phantom{\rule{0.2em}{0ex}}\frac{22}{35}$	48. $-\phantom{\rule{0.2em}{0ex}}\frac{33}{49}-\phantom{\rule{0.2em}{0ex}}\frac{18}{35}$
49. $-\phantom{\rule{0.2em}{0ex}}\frac{2}{3}-\left(-\phantom{\rule{0.2em}{0ex}}\frac{3}{4}\right)$	50. $-\phantom{\rule{0.2em}{0ex}}\frac{3}{4}-\left(-\phantom{\rule{0.2em}{0ex}}\frac{4}{5}\right)$
51. $1+\frac{7}{8}$	52. $1-\phantom{\rule{0.2em}{0ex}}\frac{3}{10}$
53. $\frac{x}{3}+\frac{1}{4}$	54. $\frac{y}{2}+\frac{2}{3}$
55. $\frac{y}{4}-\phantom{\rule{0.2em}{0ex}}\frac{3}{5}$	56. $\frac{x}{5}-\phantom{\rule{0.2em}{0ex}}\frac{1}{4}$

Mixed Practice

In the following exercises, simplify.

57. a) $\frac{2}{3}+\frac{1}{6}$ b) $\frac{2}{3}\div \frac{1}{6}$	58. a) $-\phantom{\rule{0.2em}{0ex}}\frac{2}{5}-\phantom{\rule{0.2em}{0ex}}\frac{1}{8}$ b) $-\phantom{\rule{0.2em}{0ex}}\frac{2}{5}\cdot \frac{1}{8}$
59. a) $\frac{5n}{6}\div \frac{8}{15}$ b) $\frac{5n}{6}-\phantom{\rule{0.2em}{0ex}}\frac{8}{15}$	60. a) $\frac{3a}{8}\div \frac{7}{12}$ b) $\frac{3a}{8}-\phantom{\rule{0.2em}{0ex}}\frac{7}{12}$
61. $-\phantom{\rule{0.2em}{0ex}}\frac{3}{8}\div \left(-\phantom{\rule{0.2em}{0ex}}\frac{3}{10}\right)$	62. $-\phantom{\rule{0.2em}{0ex}}\frac{5}{12}\div \left(-\phantom{\rule{0.2em}{0ex}}\frac{5}{9}\right)$
63. $-\phantom{\rule{0.2em}{0ex}}\frac{3}{8}+\frac{5}{12}$	64. $-\phantom{\rule{0.2em}{0ex}}\frac{1}{8}+\frac{7}{12}$
65. $\frac{5}{6}-\phantom{\rule{0.2em}{0ex}}\frac{1}{9}$	66. $\frac{5}{9}-\phantom{\rule{0.2em}{0ex}}\frac{1}{6}$
67. $-\phantom{\rule{0.2em}{0ex}}\frac{7}{15}-\phantom{\rule{0.2em}{0ex}}\frac{y}{4}$	68. $-\phantom{\rule{0.2em}{0ex}}\frac{3}{8}-\phantom{\rule{0.2em}{0ex}}\frac{x}{11}$
69. $\frac{11}{12a}\cdot \frac{9a}{16}$	70. $\frac{10y}{13}\cdot \frac{8}{15y}$

Use the Order of Operations to Simplify Complex Fractions

In the following exercises, simplify.

71. $\frac{{2}^{3}+{4}^{2}}{{\left(\frac{2}{3}\right)}^{2}}$	72. $\frac{{3}^{3}-{3}^{2}}{{\left(\frac{3}{4}\right)}^{2}}$
73. $\frac{{\left(\frac{3}{5}\right)}^{2}}{{\left(\frac{3}{7}\right)}^{2}}$	74. $\frac{{\left(\frac{3}{4}\right)}^{2}}{{\left(\frac{5}{8}\right)}^{2}}$
75. $\frac{2}{\frac{1}{3}+\frac{1}{5}}$	76. $\frac{5}{\frac{1}{4}+\frac{1}{3}}$
77. $\frac{\frac{7}{8}-\phantom{\rule{0.2em}{0ex}}\frac{2}{3}}{\frac{1}{2}+\frac{3}{8}}$	78. $\frac{\frac{3}{4}-\phantom{\rule{0.2em}{0ex}}\frac{3}{5}}{\frac{1}{4}+\frac{2}{5}}$
79. $\frac{1}{2}+\frac{2}{3}\cdot \frac{5}{12}$	80. $\frac{1}{3}+\frac{2}{5}\cdot \frac{3}{4}$
81. $1-\phantom{\rule{0.2em}{0ex}}\frac{3}{5}\div \frac{1}{10}$	82. $1-\phantom{\rule{0.2em}{0ex}}\frac{5}{6}\div \frac{1}{12}$
83. $\frac{2}{3}+\frac{1}{6}+\frac{3}{4}$	84. $\frac{2}{3}+\frac{1}{4}+\frac{3}{5}$
85. $\frac{3}{8}-\phantom{\rule{0.2em}{0ex}}\frac{1}{6}+\frac{3}{4}$	86. $\frac{2}{5}+\frac{5}{8}-\phantom{\rule{0.2em}{0ex}}\frac{3}{4}$
87. $12\left(\frac{9}{20}-\phantom{\rule{0.2em}{0ex}}\frac{4}{15}\right)$	88. $8\left(\frac{15}{16}-\phantom{\rule{0.2em}{0ex}}\frac{5}{6}\right)$
89. $\frac{\frac{5}{8}+\frac{1}{6}}{\frac{19}{24}}$	90. $\frac{\frac{1}{6}+\frac{3}{10}}{\frac{14}{30}}$
91. $\left(\frac{5}{9}+\frac{1}{6}\right)\div \left(\frac{2}{3}-\phantom{\rule{0.2em}{0ex}}\frac{1}{2}\right)$	92. $\left(\frac{3}{4}+\frac{1}{6}\right)\div \left(\frac{5}{8}-\phantom{\rule{0.2em}{0ex}}\frac{1}{3}\right)$

Evaluate Variable Expressions with Fractions

In the following exercises, evaluate.

93. $x+\left(-\phantom{\rule{0.2em}{0ex}}\frac{5}{6}\right)$ when a) $x=\frac{1}{3}$ b) $x=-\phantom{\rule{0.2em}{0ex}}\frac{1}{6}$	94. $x+\left(-\phantom{\rule{0.2em}{0ex}}\frac{11}{12}\right)$ when a) $x=\frac{11}{12}$ b) $x=\frac{3}{4}$
95. $x-\phantom{\rule{0.2em}{0ex}}\frac{2}{5}$ when a) $x=\frac{3}{5}$ b) $x=-\phantom{\rule{0.2em}{0ex}}\frac{3}{5}$	96. $x-\phantom{\rule{0.2em}{0ex}}\frac{1}{3}$ when a) $x=\frac{2}{3}$ b) $x=-\phantom{\rule{0.2em}{0ex}}\frac{2}{3}$
97. $\frac{7}{10}-w$ when a) $w=\frac{1}{2}$ b) $w=-\phantom{\rule{0.2em}{0ex}}\frac{1}{2}$	98. $\frac{5}{12}-w$ when a) $w=\frac{1}{4}$ b) $w=-\phantom{\rule{0.2em}{0ex}}\frac{1}{4}$
99. $2{x}^{2}{y}^{3}$ when $x=-\phantom{\rule{0.2em}{0ex}}\frac{2}{3}$ and $y=-\phantom{\rule{0.2em}{0ex}}\frac{1}{2}$	100. $8{u}^{2}{v}^{3}$ when $u=-\phantom{\rule{0.2em}{0ex}}\frac{3}{4}$ and $v=-\phantom{\rule{0.2em}{0ex}}\frac{1}{2}$
101. $\frac{a+b}{a-b}$ when $a=-3,b=8$	102. $\frac{r-s}{r+s}$ when $r=10,s=-5$

Everyday Math

103. Decorating Laronda is making covers for the throw pillows on her sofa. For each pillow cover, she needs $\frac{1}{2}$ yard of print fabric and $\frac{3}{8}$ yard of solid fabric. What is the total amount of fabric Laronda needs for each pillow cover?

104. Baking Samuel is baking chocolate chip cookies and oatmeal cookies. He needs $\frac{1}{2}$ cup of sugar for the chocolate chip cookies and $\frac{1}{4}$ of sugar for the oatmeal cookies. How much sugar does he need altogether?

Answers

1. $\frac{11}{13}$	3. $\frac{x+3}{4}$	5. $-\phantom{\rule{0.2em}{0ex}}\frac{5}{8}$
7. $\frac{7}{17}$	9. $-\phantom{\rule{0.2em}{0ex}}\frac{16}{13}$	11. $\frac{4}{15}$
13. $\frac{1}{2}$	15. $\frac{5}{7}$	17. $\frac{5y-7}{8}$
19. $-\phantom{\rule{0.2em}{0ex}}\frac{38}{u}$	21. $\frac{1}{5}$	23. $-\phantom{\rule{0.2em}{0ex}}\frac{2}{9}$
25. $-\phantom{\rule{0.2em}{0ex}}\frac{1}{4}$	27. $\frac{n-4}{5}$	29. $-\phantom{\rule{0.2em}{0ex}}\frac{5}{24}$
31. $\frac{2}{9}$	33. $\frac{9}{14}$	35. $\frac{4}{9}$
37. $\frac{29}{24}$	39. $\frac{1}{48}$	41. $\frac{7}{24}$
43. $\frac{37}{120}$	45. $\frac{17}{105}$	47. $-\phantom{\rule{0.2em}{0ex}}\frac{53}{40}$
49. $\frac{1}{12}$	51. $\frac{15}{8}$	53. $\frac{4x+3}{12}$
55. $\frac{4y-12}{20}$	57. a) $\frac{5}{6}$ b) 4	59. a) $\frac{25n}{16}$ b) $\frac{25n-16}{30}$
61. $\frac{5}{4}$	63. $\frac{1}{24}$	65. $\frac{13}{18}$
67. $\frac{-28-15y}{60}$	69. $\frac{33}{64}$	71. 54
73. $\frac{49}{25}$	75. $\frac{15}{4}$	77. $\frac{5}{21}$
79. $\frac{7}{9}$	81. $-5$	83. $\frac{19}{12}$
85. $\frac{23}{24}$	87. $\frac{11}{5}$	89. 1
91. $\frac{13}{3}$	93. a) $-\phantom{\rule{0.2em}{0ex}}\frac{1}{2}$ b) $-1$	95. a) $\frac{1}{5}$ b) $-1$
97. a) $\frac{1}{5}$ b) $\frac{6}{5}$	99. $-\phantom{\rule{0.2em}{0ex}}\frac{1}{9}$	101. $-\phantom{\rule{0.2em}{0ex}}\frac{5}{11}$
103. $\frac{7}{8}$ yard

Attributions

This chapter has been adapted from “Add and Subtract Fractions” in Elementary Algebra (OpenStax) by Lynn Marecek and MaryAnne Anthony-Smith, which is under a CC BY 4.0 Licence. Adapted by Izabela Mazur. See the Copyright page for more information.

License

Icon for the Creative Commons Attribution 4.0 International License

To evaluate $x+\dfrac{1}{3}$ when $x=-\phantom{\rule{0.2em}{0ex}}\dfrac{1}{3}$ , substitute $-\phantom{\rule{0.2em}{0ex}}\dfrac{1}{3}$ for $x$ in the expression.	$x+\dfrac{1}{3}$
Substitute $\color{red}-\dfrac{1}{3}$ for $x$ .	${\color{red}-\dfrac{1}{3}}+\dfrac{1}{3}$
Simplify. $\phantom{\rule{18em}{0ex}}$	$0$

To evaluate $x+\dfrac{1}{3}$ when $x=-\phantom{\rule{0.2em}{0ex}}\dfrac{3}{4}$ , we substitute $-\phantom{\rule{0.2em}{0ex}}\dfrac{3}{4}$ for x in the expression.	$x+\dfrac{1}{3}$
Substitute $\color{red}-\dfrac{3}{4}$ for $x$ .	${\color{red}-\dfrac{3}{4}}+\dfrac{1}{3}$
Rewrite as equivalent fractions with the LCD, 12.	$-\dfrac{3 \cdot {\color{red}3}}{4 \cdot {\color{red}3}}+\dfrac{1 \cdot {\color{red}4}}{3 \cdot {\color{red}4}}$
Simplify.	$-\dfrac{9}{12}+\dfrac{4}{12}$
Add.	$-\phantom{\rule{0.2em}{0ex}}\dfrac{5}{12}$

Add or Subtract Fractions with a Common Denominator

Add or Subtract Fractions with Different Denominators

Use the Order of Operations to Simplify Complex Fractions

Evaluate Variable Expressions with Fractions

Key Concepts

Glossary

Practice Makes Perfect

Add and Subtract Fractions with a Common Denominator

Mixed Practice

Add or Subtract Fractions with Different Denominators

Mixed Practice

Use the Order of Operations to Simplify Complex Fractions

Evaluate Variable Expressions with Fractions

Everyday Math

Answers

Attributions

This chapter has been adapted from “Add and Subtract Fractions” in Elementary Algebra (OpenStax) by Lynn Marecek and MaryAnne Anthony-Smith, which is under a CC BY 4.0 Licence. Adapted by Izabela Mazur. See the Copyright page for more information.

License

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