7.5: Addition and Subtraction of Rational Functions

Learning Outcomes

Perform addition and subtraction of rational expressions
Perform addition and subtraction of rational functions

The concept and skills for adding or subtracting rational expressions are similar to those for adding or subtracting numerical fractions. We will find a common denominator (or common multiple) among the rational expressions, build each rational expression to get all rational expressions with the same denominator, and then perform the addition or subtraction on the numerators. The common denominator is the common multiple of the denominators. Although any common multiple of the denominators can be used as the common denominator for adding or subtracting the rational expressions, the least common multiple is recommended because it could help us avoid complicated algebra or further simplification of our answer. Therefore, we use the Least Common Multiple (LCM) as the common denominator when adding or subtracting fractions.

Least Common Multiple

To find the least common multiple (LCM) of two or more algebraic expressions, we factor the expressions and multiply all of the distinct factors.

Example 1

Determine the least common multiple of $x^2-4,\;x^2-4x-5$ and $x^2-7x+10$ .

Solution

We factor each expression:

$x^2-4=\color{blue}{(x-2)}(x+2)$

$x^2-4x-5=\color{red}{(x-5)}(x+1)$

$x^2-7x+10=\color{red}{(x-5)}\color{blue}{(x-2)}$

The LCM must contain the highest power of every factor:

LCM $=\color{blue}{(x-2)}(x+2)\color{red}{(x-5)}(x+1)$

The colors indicate factors that are common to two expressions. Notice that they are written in the LCM only once.

Try It 1

Determine the least common multiple of $x^2-9,\;x^2-x-6$ and $x^2-6x+9$ .

Show Answer

LCM $=(x-3)^2(x+2)(x+3)$

Adding and Subtracting Rational Expressions

To add rational expressions we need a common denominator. We can use the LCM of the denominators as the common denominator.

Example 2

Add: $\dfrac{5x-3}{x}+\dfrac{6x+4}{x}$ . State any restrictions on the variable.

Solution

In this example we already have a common denominator, $x$ .

$\begin{aligned}\dfrac{5x-3}{x}+\dfrac{6x+4}{x}&=\dfrac{5x-3+6x+4}{x}&&\text{Combine numerators with the common denominator. }x\neq 0\\\\&=\dfrac{11x+1}{x}&&\text{Simplify numerator}\end{aligned}$

Try It 2

Add: $\dfrac{4x-1}{x+2}+\dfrac{-3x+4}{x+2}$ . State any restrictions on the variable.

Show Answer

\frac{x + 3}{x + 2} x \neq - 2

$\dfrac{x+3}{x+2}\;\;\;\;x\neq -2$

To subtract rational expressions, we need to subtract the complete numerator of the second fraction.

Example 3

Subtract: $\dfrac{4x}{x-1}-\dfrac{x-5}{x-1}$

Solution

In this example we have a common denominator, $x-1$ , with a restriction that $x\neq 1$ .

$\begin{aligned}\dfrac{4x}{x-1}-\dfrac{x-5}{x-1}&=\dfrac{4x-(x-5)}{x-1}&&\text{Combine the numerators}\\\\&=\dfrac{4x-x+5}{x-1}&&\text{Subtract using the distributive property}\\\\&=\dfrac{3x+5}{x-1}&&\text{Simplify}\end{aligned}$

Try It 3

Subtract: $\dfrac{4x-3}{x+5}-\dfrac{2x-5}{x+5}$

Show Answer

\frac{2 x + 2}{x + 5}

$\dfrac{2x+2}{x+5}$

Usually, we need to start by finding a common denominator and building equivalent fractions with that new denominator. To build a fraction, we multiply the numerator and denominator by the same factors.

Example 4

Add: $\dfrac{2x^2}{x^2-4}+\dfrac{x}{x-2}$ . State any restrictions on the variable.

Solution

$\begin{aligned}\dfrac{2x^2}{x^2-4}+\dfrac{x}{x-2}&=\dfrac{2x^2}{(x-2)(x+2)}+\dfrac{x}{(x-2)}&&\text{Factor. LCM }=(x-2)(x+2)\;\;x\neq\pm2\\\\&=\dfrac{2x^2}{(x-2)(x+2)}+\dfrac{x\color{blue}{(x+2)}}{(x-2)\color{blue}{(x+2)}}&&\text{Build fractions with a common denominator}\\\\&=\dfrac{2x^2+x(x+2)}{(x-2)(x+2)}&&\text{Combine numerators; use common denominator}\\\\&=\dfrac{2x^2+x^2+2x}{(x-2)(x+2)}&&\text{Distribute }x\text{ into }(x+2)\\\\&=\dfrac{3x^2+2x}{(x-2)(x+2)}&&\text{Simplify}\\\\&=\dfrac{x(3x+2)}{(x-2)(x+2)}&&\text{Factor numerator to see if fraction will simplify}\end{aligned}$

Try It 4

Add: $\dfrac{4x+3}{x^2-25}+\dfrac{5x-1}{x+5}$ . State any restrictions on the variable.

Show Answer

$\dfrac{5x^2-22x+8}{(x-5)(x+5)}=\dfrac{(x-4)(5x-2)}{(x-5)(x+5)}\;\;x\neq\pm5$

Example 5

Subtract: $\dfrac{2x^2}{x^2-4x-5}-\dfrac{3x}{x-5}$ . State any restrictions on the variable.

Solution

$\begin{aligned}\dfrac{2x^2}{x^2-4x-5}-\dfrac{3x}{x-5}&=\dfrac{2x^2}{(x-5)(x+1)}-\dfrac{3x}{(x-5)}&&\text{Factor. LCM }=(x-5)(x+1)\;\;x\neq5,\,-1\\\\&=\dfrac{2x^2}{(x-5)(x+1)}-\dfrac{3x\color{blue}{(x+1)}}{(x-5)\color{blue}{(x+1)}}&&\text{Build fractions using LCM as common denominator}\\\\&=\dfrac{2x^2-3x(x+1)}{(x-5)(x+1)}&&\text{Combine fractions}\\\\&=\dfrac{2x^2-3x^2-3x}{(x-5)(x+1)}&&\text{Distribute }-3x\text{ into }(x+1)\\\\&=\dfrac{-x^2-3x}{(x-5)(x+1)}&&\text{Simplify}\\\\&=\dfrac{-x(x+3)}{(x-5)(x+1)}&&\text{Factor to look for common factors}\end{aligned}$

Try It 5

Subtract: $\dfrac{4x+3}{x^2+9x+8}-\dfrac{2x-3}{x+1}$ . State any restrictions on the variable.

Show Answer

$\dfrac{-2x^2-9x+27}{(x+1)(x+8)}\;\;x\neq-8,\,-1$

The video presents an example of adding two rational expressions whose denominators are binomials with no common factors.

The next video contains an example of subtracting rational expressions.

Adding and Subtracting Rational Functions

Now, we may use the methods we learned above to add two rational functions. The domain of the sum function will equal all real numbers with the combined restrictions of each function.

adding and subtracting functions

$\left(f+g\right)(x)=f(x)+g(x)$

$\left(f-g\right)(x)=f(x)-g(x)$

The domain of the sum or difference function will be all real numbers except the combined restricted values of $f(x)$ and $g(x)$ .

Example 6

Add the two functions $f(x)=\dfrac{x+2}{x^2-1}$ and $g(x)=\dfrac{x-3}{x^2+x-2}$ . State the domain of the function. State the restrictions and the domain.

Solution

Before we perform the addition, we need to discuss the domain. The denominators of the two functions cannot be zero because it is not defined for a fraction with zero in the denominator. The denominator $x^2-1$ can be factored as $(x-1)(x+1)$ . Setting $(x-1)(x+1)=0$ results in $x=1, -1$ . Therefore, $x \neq 1, -1$ . The denominator $x^2+x-2$ can be factored as $(x+2)(x-1)$ . Setting $(x+2)(x-1)=0$ results in $x=-2, 1$ . Therefore, $x \neq -2, 1$ . In summary, the domain of $(f+g)(x)$ is all real numbers except for the numbers $1, -1, -2$ .

Domain = $\{x\;|\;x\in\mathbb{R},\;x\neq\pm1,\,-2\}$

$\begin{aligned}(f+g)(x)&=\dfrac{x+2}{x^2-1}+\dfrac{x-3}{x^2+x-2}&&\text{Factor}\\\\&=\dfrac{x+2}{(x-1)(x+1)}+\dfrac{x-3}{(x+2)(x-1)}&&\text{LCD }=(x-1)(x+1)(x+2)\\\\&=\dfrac{x+2}{(x-1)(x+1)}\color{blue}{\times\frac{(x+2)}{(x+2)}}+\dfrac{x-3}{(x+2)(x-1)}\color{blue}{\times \frac{(x+1)}{(x+1)}}&&\text{Build equivalent fractions}\\\\&=\dfrac{x^2+4x+4}{(x-1)(x+1)(x+2)}+\dfrac{x^2-2x-3}{(x-1)(x+1)(x+2)}&&\text{Combine numerators}\\\\&=\dfrac{(x^2+4x+4)+(x^2-2x-3)}{(x+1)(x-1)(x+2)}&&\text{Add like terms}\\\\&=\dfrac{2x^2+2x+1}{(x+1)(x-1)(x+2)}\end{aligned}$

Try It 6

Add the two functions $f(x)=\dfrac{x}{(x-3)^2}$ and $g(x)=\dfrac{x-1}{x^2-2x-3}$ . State the domain of the function.

Show Answer

$f(x)+g(x)=\dfrac{2x^2-3x+3}{(x-3)^2(x+1)}\;\;\;\;x\neq3,\,-1$

Domain of $\left(f+g\right)(x)=\{x\;|\;\in\mathbb{R},\;x\neq3,\,-1\}$

Example 7

Subtract the function $g(x)=\dfrac{x+3}{3x^2+9x+6}$ from the function $f(x)=\dfrac{x+2}{x^2-2x-8}$ . State the restrictions and the domain.

Solution

Before we perform the addition, we need to discuss the domain first. The denominators of the two functions (or rational expressions) cannot be zero because it is not defined for a fraction with zero in the denominator. The denominator $x^2-2x-8$ can be factored as $(x-4)(x+2)$ . Let $(x-4)(x+2)=0$ . According to the zero product rule, $x=4, -2$ . Therefore, $x \neq 4, -2$ . The denominator $3x^2+9x+6$ can be factored as $3(x+1)(x+2)$ . Let $3(x+1)(x+2)=0$ . According to the zero product rule, $x=-1, -2$ . Therefore, $x \neq -1, -2$ . In summary, the domain of $(f-g)(x)$ is all real numbers except for the numbers $4, -1, -2$ .

$\begin{aligned}(f-g)(x)&=\dfrac{x+2}{x^2-2x-8}-\dfrac{x+3}{3x^2+9x+6}\\\\&=\dfrac{x+2}{(x-4)(x+2)}-\dfrac{x+3}{3(x^2+3x+2)}&&\text{Factor}\\\\&=\dfrac{x+2}{(x-4)(x+2)}-\dfrac{x+3}{3(x+1)(x+2)}&&\text{LCD }=3(x+1)(x+2)(x-4)\\\\&=\dfrac{x+2}{(x-4)(x+2)} \times \color{blue}{\frac{3(x+1)}{3(x+1)}}-\dfrac{x+3}{3(x+1)(x+2)} \times \color{blue}{\frac{(x-4)}{(x-4)}}&&\text{Build equivalent fractions}\\\\&=\dfrac{3x^2+9x+6}{3(x-4)(x+2)(x+1)}-\dfrac{x^2-x-12}{3(x-4)(x+2)(x+1)}&&\text{Combine numerators}\\\\&=\dfrac{3x^2+9x+6-(x^2-x-12)}{3(x-4)(x+2)(x+1)}&&\text{Distribute the }-\text{ sign}\\\\&=\dfrac{3x^2+9x+6-x^2+x+12}{3(x-4)(x+2)(x+1)}&&\text{Combine like terms}\\&=\dfrac{2x^2+10x+18}{3(x-4)(x+2)(x+1)}&&\text{Factor}\\\\&=\dfrac{2(x^2+5x+9)}{3(x-4)(x+2)(x+1)}\end{aligned}$

Try It 7

Subtract the function $g(x)=\dfrac{2}{x^2-4}$ from the function $f(x)=\dfrac{6}{x^2+4x+4}$ .

Show Answer

$f(x)-g(x)=\dfrac{4(x-4)}{(x+2)^2(x-2)}$

Domain of $\left(f-g\right)(x)=\{x\;|\;x\in\mathbb{R},\;x\neq\pm2\}$

Adding and subtracting rational functions can be extended to include more than two functions.

Example 8

Combine $f(x)+g(x)-h(x)$ , when $f(x)=\dfrac{2x}{x-1},\;g(x)=\dfrac{1}{x},\;h(x)=\dfrac{2x-1}{x^2-x}$ . State the domain of $\left(f+g-h\right)(x)$ .

Solution

$\begin{aligned}f(x)+g(x)-h(x)&=\dfrac{2x}{x-1}+\dfrac{1}{x}-\dfrac{2x-1}{x^2-x}\\\\&=\dfrac{2x}{(x-1)}+\dfrac{1}{x}-\dfrac{(2x-1)}{x(x-1)}&&\text{Factor. LCM}=x(x-1),\;x\neq0,\,1\\\\&=\color{blue}{\dfrac{x}{x}}\cdot\dfrac{2x}{(x-1)}+\color{blue}{\dfrac{(x-1)}{(x-1)}}\cdot\dfrac{1}{x}-\dfrac{(2x-1)}{x(x-1)}&&\text{Build fractions with common denominator}\\\\&=\dfrac{2x^2+(x-1)-(2x-1)}{x(x-1)}&&\text{Combine numerators}\\\\&=\dfrac{2x^2+x-1-2x+1}{x(x-1)}&&\text{distribute + and – signs}\\\\&=\dfrac{2x^2-x}{x(x-1)}&&\text{Combine like terms}\\\\&=\dfrac{x(2x-1)}{x(x-1)}&&\text{Factor numerator}\\\\&=\dfrac{2x-1}{x-1}&&\text{Cancel: }\dfrac{x}{x}=1\end{aligned}$

Domain of $\left(f+g-h\right)(x)=\{x\;|\;x\in\mathbb{R},\;x\neq0,\,1\}$

Try It 8

Combine $f(x)+g(x)-h(x)$ , when $f(x)=\dfrac{x}{x+1},\;g(x)=\dfrac{3}{x-1},\;h(x)=\dfrac{6}{x^2-1}$ . State the domain of $\left(f+g-h\right)(x)$ .

Show Answer

$\left(f+g-h\right)(x)=\dfrac{x+3}{x+1}$

Domain of $\left(f+g-h\right)(x)=\{x\;|\;x\in\mathbb{R},\;x\neq\pm1\}$

Try It 9

Combine $f(x)+g(x)-h(x)$ , when $f(x)=\dfrac{2x}{x+5},\;g(x)=\dfrac{3}{x-3},\;h(x)=\dfrac{13x+15}{x^2+2x-15}$ . State the domain of $\left(f+g-h\right)(x)$ .

Show Answer

$\left(f+g-h\right)(x)=\dfrac{2x(x-8)}{(x-3)(x+5)}$

Domain of $\left(f+g-h\right)(x)=\{x\;|\;\mathbb{R},\;x\neq -5,\;3\}$

Chapter 7: Rational Functions