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 ratio 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.
=4515=415=35\begin{aligned} &\phantom{=}\dfrac{\blueD4}{\purpleC5}-\dfrac{\blueD1}{\purpleC5}\\\\\\ &=\dfrac{\blueD{4}-\blueD{1}}{\purpleC 5}\\ \\ &=\dfrac{3}{5} \end{aligned}

Variable expressions

The process is the same with rational expressions:
=7a+3a+2+2a1a+2=(7a+3)+(2a1)a+2Add=7a+3+2a1a+2Remove parentheses=9a+2a+2Combine like terms\begin{aligned} &\phantom{=}\dfrac{\blueD{7a+3}}{\purpleC{a+2}}+\dfrac{\blueD{2a-1}}{\purpleC{a+2}}\\\\\\ &=\dfrac{(\blueD{7a+3})+(\blueD{2a-1})}{\purpleC{a+2}}&&\small{\gray{\text{Add}}}\\ \\ &=\dfrac{{7a+3}+{2a-1}}{{a+2}}&&\small{\gray{\text{Remove parentheses}}}\\ \\ &=\dfrac{9a+2}{a+2}&&\small{\gray{\text{Combine like terms}}} \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:
=b+1b24bb2=(b+1)(4b)b2Subtract=b+14+bb2Remove parentheses and distribute=2b3b2Combine like terms\begin{aligned} &\phantom{=}\dfrac{\blueD{b+1}}{\purpleC{b^2}}-\dfrac{\blueD{4-b}}{\purpleC{b^2}}\\\\\\ &=\dfrac{(\blueD{b+1})-(\blueD{4-b})}{\purpleC{b^2}}&&\small{\gray{\text{Subtract}}}\\ \\ &=\dfrac{b+1-4+b}{{b^2}}&&\small{\gray{\text{Remove parentheses and distribute}}}\\ \\ &=\dfrac{2b-3}{b^2}&&\small{\gray{\text{Combine like terms}}} \end{aligned}

Check your understanding

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

=x+5x1+2x3x1=(x+5)+(2x3)x1Add=x+5+2x3x1Remove parentheses=3x+2x1Combine like terms\begin{aligned} &\phantom{=}\dfrac{x+5}{x-1}+\dfrac{2x-3}{x-1}\\\\\\ &=\dfrac{(x+5)+(2x-3)}{x-1}&&\small{\gray{\text{Add}}}\\ \\ &=\dfrac{x+5+2x-3}{x-1}&&\small{\gray{\text{Remove parentheses}}}\\ \\ &=\dfrac{3x+2}{x-1}&&\small{\gray{\text{Combine like terms}}}\\ \end{aligned}
2) 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

=x+12x5x22x=(x+1)(5x2)2xSubtract=x+15x+22xRemove parentheses and distribute=4x+32xCombine like terms\begin{aligned} &\phantom{=}\dfrac{x+1}{2x}-\dfrac{5x-2}{2x}\\\\\\ &=\dfrac{(x+1)-(5x-2)}{2x}&&\small{\gray{\text{Subtract}}}\\ \\ &=\dfrac{x+1-5x+2}{2x}&&\small{\gray{\text{Remove parentheses and distribute}}}\\ \\ &=\dfrac{-4x+3}{2x}&&\small{\gray{\text{Combine like terms}}}\\ \end{aligned}

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.
=23+12=23(22)+12(33)Create common denominators=46+36=76\begin{aligned} &\phantom{=}\dfrac{2}{\blueD3}+\dfrac{1}{\tealD2}\\\\\\ &=\dfrac{2}{\blueD3} \left(\tealD{\dfrac{2}{2}}\right)+\dfrac{1}{\tealD2}\left( \blueD{\dfrac{3}{3}}\right)&&\small{\gray{\text{Create common denominators}}}\\ \\ &=\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 left parenthesis, start color blueD, 3, end color blueD, right parenthesis needed a factor of start color tealD, 2, end color tealD.
  • The denominator in the second fraction left parenthesis, start color tealD, 2, end color tealD, right parenthesis needed a factor of start color blueD, 3, end color blueD.
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 blueD, x, minus, 3, end color blueD, end fraction, plus, start fraction, 2, divided by, start color tealD, x, plus, 5, end color tealD, end fraction
In order for the two denominators to be the same, the first needs a factor of start color tealD, x, plus, 5, end color tealD and the second needs a factor of start color blueD, x, minus, 3, end color blueD. Let's manipulate the fractions in order to achieve this. Then, we can add as usual.
=1x3+2x+5=1x3(x+5x+5)+2x+5(x3x3)Create common denominators=1(x+5)(x3)(x+5)+2(x3)(x+5)(x3)=1(x+5)+2(x3)(x3)(x+5)Add=1x+5+2x6(x3)(x+5)=3x1(x3)(x+5)\begin{aligned} &\phantom{=}{\dfrac{1}{\blueD{x-3}}+\dfrac{2}{\tealD{x+5}}}\\\\\\ &=\dfrac{1}{\blueD{x-3}}{\left(\tealD{\dfrac{x+5}{x+5}}\right)}+\dfrac{2}{\tealD{x+5}}{\left(\blueD{\dfrac{x-3}{x-3}}\right)}&&\small{\gray{\text{Create common denominators}}}\\\\\\ &=\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)}&&\small{\gray{\text{Add}}}\\ \\\\\\ &=\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!
Technically speaking, start fraction, x, minus, 3, divided by, x, minus, 3, end fraction, equals, 1 for x, does not equal, 3 and start fraction, x, plus, 5, divided by, x, plus, 5, end fraction, equals, 1 for x, does not equal, minus, 5.
However, in order for start fraction, 1, divided by, start color blueD, x, minus, 3, end color blueD, end fraction, plus, start fraction, 2, divided by, start color tealD, x, plus, 5, end color tealD, end fraction to be defined, we must require start color blueD, x, does not equal, 3, end color blueD and start color tealD, x, does not equal, minus, 5, end color tealD. Therefore, in this example, we can be confident that start fraction, x, minus, 3, divided by, x, minus, 3, end fraction, equals, 1 and start fraction, x, plus, 5, divided by, x, plus, 5, end fraction, equals, 1.
In the last two steps, we simplified the numerator. While you can also multiply left parenthesis, x, minus, 3, right parenthesis and left parenthesis, x, plus, 5, right parenthesis in the denominator, it is common to leave this in factored form.

Check your understanding

3) start fraction, 3, divided by, x, plus, 4, end fraction, plus, start fraction, 2, divided by, x, minus, 2, end fraction, equals

In order for the two denominators to be the same, the first fraction needs a factor of x, minus, 2 and the second fraction needs a factor of x, plus, 4. Let's manipulate the fractions in order to achieve this. Then, we can add as usual.
=3x+4+2x2=3x+4(x2x2)+2x2(x+4x+4)Create common denominators=3(x2)(x+4)(x2)+2(x+4)(x2)(x+4)=3(x2)+2(x+4)(x2)(x+4)Add=3x6+2x+8(x2)(x+4)=5x+2(x2)(x+4)\begin{aligned} &\phantom{=}\dfrac{3}{x+4}+\dfrac{2}{x-2}\\\\\\ &=\dfrac3{x+4}\left(\dfrac{x-2}{x-2}\right)+\dfrac{2}{x-2}\left(\dfrac{x+4}{x+4}\right)&&\small{\gray{\text{Create common denominators}}}\\ \\\\\\ &=\dfrac{3(x-2)}{(x+4)(x-2)}+\dfrac{2(x+4)}{(x-2)(x+4)}\\ \\\\\\ &=\dfrac{3(x-2)+2(x+4)}{(x-2)(x+4)}&&\small{\gray{\text{Add}}}\\ \\\\\\ &=\dfrac{3x-6+2x+8}{(x-2)(x+4)}\\ \\\\\\ &=\dfrac{5x+2}{(x-2)(x+4)} \end{aligned}
4) start fraction, 2, divided by, x, minus, 1, end fraction, minus, start fraction, 5, divided by, x, end fraction, equals

In order for the two denominators to be the same, the first fraction needs a factor of x and the second fraction needs a factor of x, minus, 1. Let's manipulate the fractions in order to achieve this. Then, we can subtract as usual.
=2x15x=2x1(xx)5x(x1x1)Create common denominators=2xx(x1)5(x1)x(x1)=2x5(x1)x(x1)Subtract=2x5x+5x(x1)=3x+5x(x1)\begin{aligned} &\phantom{=}\dfrac{2}{x-1}-\dfrac{5}{x}\\\\\\ &=\dfrac2{x-1}\left(\dfrac{x}{x}\right)-\dfrac{5}{x}\left(\dfrac{x-1}{x-1}\right)&&\small{\gray{\text{Create common denominators}}}\\ \\\\\\ &=\dfrac{2x}{x(x-1)}-\dfrac{5(x-1)}{x(x-1)}\\ \\\\\\ &=\dfrac{2x-5(x-1)}{x(x-1)}&&\small{\gray{\text{Subtract}}}\\ \\\\\\ &=\dfrac{2x-5x+5}{x(x-1)}\\ \\\\\\ &=\dfrac{-3x+5}{x(x-1)} \end{aligned}

What's next?

Our next article covers more challenging examples of adding and subtracting rational expressions.
You will learn about the least common denominator, and why it is important to use this as the common denominator when adding or subtracting rational expressions.