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# Intro to adding & subtracting rational expressions

Learn how to add or subtract two rational expressions into a single expression.

#### What you should be familiar with before taking this lesson

A rational expression is a quotient of two polynomials. For example, the expression start fraction, x, plus, 2, divided by, x, plus, 1, end fraction is a rational expression.
If you are unfamiliar with rational expressions, you may want to check out our intro to rational expressions.

#### What you will learn in this lesson

In this lesson, you will learn how to add and subtract rational expressions.

## Adding and subtracting rational expressions (common denominators)

### Numerical fractions

We can add and subtract rational expressions in much the same way as we add and subtract numerical fractions.
To add or subtract two numerical fractions with the same denominator, we simply add or subtract the numerators, and write the result over the common denominator.
\begin{aligned} &\phantom{=}\dfrac{\blueE4}{\purpleD5}-\dfrac{\blueE1}{\purpleD5} \\\\ &=\dfrac{\blueE{4}-\blueE{1}}{\purpleD 5} \\\\ &=\dfrac{3}{5} \end{aligned}

### Variable expressions

The process is the same with rational expressions:
\begin{aligned} &\phantom{=}\dfrac{\blueE{7a+3}}{\purpleD{a+2}}+\dfrac{\blueE{2a-1}}{\purpleD{a+2}} \\\\ &=\dfrac{(\blueE{7a+3})+(\blueE{2a-1})}{\purpleD{a+2}} \\\\ &=\dfrac{{7a+3}+{2a-1}}{{a+2}} \\\\ &=\dfrac{9a+2}{a+2} \end{aligned}
It is good practice to place the numerators in parentheses, especially when subtracting rational expressions. This way, we are reminded to distribute the negative sign!
For example:
\begin{aligned} &\phantom{=}\dfrac{\blueE{b+1}}{\purpleD{b^2}}-\dfrac{\blueE{4-b}}{\purpleD{b^2}} \\\\ &=\dfrac{(\blueE{b+1})-(\blueE{4-b})}{\purpleD{b^2}} \\\\ &=\dfrac{b+1-4+b}{{b^2}} \\\\ &=\dfrac{2b-3}{b^2} \end{aligned}

Problem 1
start fraction, x, plus, 5, divided by, x, minus, 1, end fraction, plus, start fraction, 2, x, minus, 3, divided by, x, minus, 1, end fraction, equals

Problem 2
Subtract.
start fraction, x, plus, 1, divided by, 2, x, end fraction, minus, start fraction, 5, x, minus, 2, divided by, 2, x, end fraction, equals

## Adding and subtracting rational expressions (different denominators)

### Numerical fractions

To understand how to add or subtract rational expressions with different denominators, let's first examine how this is done with numerical fractions.
For example, let's find start fraction, 2, divided by, 3, end fraction, plus, start fraction, 1, divided by, 2, end fraction.
\begin{aligned} &\phantom{=}\dfrac{2}{\blueE3}+\dfrac{1}{\tealE2} \\\\ &=\dfrac{2}{\blueE3} \left(\tealE{\dfrac{2}{2}}\right)+\dfrac{1}{\tealE2}\left( \blueE{\dfrac{3}{3}}\right) \\\\ &=\dfrac{4}{6}+\dfrac{3}{6} \\\\ &=\dfrac{7}{6} \end{aligned}
Notice that a common denominator of 6 was needed to add the two fractions:
• The denominator in the first fraction (start color #0c7f99, 3, end color #0c7f99) needed a factor of start color #208170, 2, end color #208170.
• The denominator in the second fraction (start color #208170, 2, end color #208170) needed a factor of start color #0c7f99, 3, end color #0c7f99.
Each fraction was multiplied by a form of 1 to obtain this.

### Variable expressions

Now let's apply this to the following example:
start fraction, 1, divided by, start color #0c7f99, x, minus, 3, end color #0c7f99, end fraction, plus, start fraction, 2, divided by, start color #208170, x, plus, 5, end color #208170, end fraction
In order for the two denominators to be the same, the first needs a factor of start color #208170, x, plus, 5, end color #208170 and the second needs a factor of start color #0c7f99, x, minus, 3, end color #0c7f99. Let's manipulate the fractions in order to achieve this. Then, we can add as usual.
\begin{aligned} &\phantom{=}{\dfrac{1}{\blueE{x-3}}+\dfrac{2}{\tealE{x+5}}} \\\\ &=\dfrac{1}{\blueE{x-3}}{\left(\tealE{\dfrac{x+5}{x+5}}\right)}+\dfrac{2}{\tealE{x+5}}{\left(\blueE{\dfrac{x-3}{x-3}}\right)} \\\\ &=\dfrac{1(x+5)}{(x-3)(x+5)}+\dfrac{2(x-3)}{(x+5)(x-3)} \\\\ &=\dfrac{1(x+5)+2(x-3)}{(x-3)(x+5)} \\\\ &=\dfrac{1x+5+2x-6}{(x-3)(x+5)} \\\\ &=\dfrac{3x-1}{(x-3)(x+5)} \end{aligned}
Notice that the first step is possible because start fraction, x, plus, 5, divided by, x, plus, 5, end fraction and start fraction, x, minus, 3, divided by, x, minus, 3, end fraction are equal to 1, and multiplication by 1 does not change the value of the expression!
In the last two steps, we rewrote the numerator. While you can also expand left parenthesis, x, minus, 3, right parenthesis, left parenthesis, x, plus, 5, right parenthesis in the denominator, it is common to leave this in factored form.

Problem 3