# 1.6 Rational expressions  (Page 2/6)

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$\frac{1}{x}\cdot \frac{3}{{x}^{2}}=\frac{3}{{x}^{3}}$

Given two rational expressions, divide them.

1. Rewrite as the first rational expression multiplied by the reciprocal of the second.
2. Factor the numerators and denominators.
3. Multiply the numerators.
4. Multiply the denominators.
5. Simplify.

## Dividing rational expressions

Divide the rational expressions and express the quotient in simplest form:

$\frac{2{x}^{2}+x-6}{{x}^{2}-1}÷\frac{{x}^{2}-4}{{x}^{2}+2x+1}$
$\frac{9{x}^{2}-16}{3{x}^{2}+17x-28}÷\frac{3{x}^{2}-2x-8}{{x}^{2}+5x-14}$

Divide the rational expressions and express the quotient in simplest form:

$\frac{9{x}^{2}-16}{3{x}^{2}+17x-28}÷\frac{3{x}^{2}-2x-8}{{x}^{2}+5x-14}$

$1$

## Adding and subtracting rational expressions

Adding and subtracting rational expressions works just like adding and subtracting numerical fractions. To add fractions, we need to find a common denominator. Let’s look at an example of fraction addition.

$\begin{array}{ccc}\hfill \frac{5}{24}+\frac{1}{40}& =& \frac{25}{120}+\frac{3}{120}\hfill \\ & =& \frac{28}{120}\hfill \\ & =& \frac{7}{30}\hfill \end{array}$

We have to rewrite the fractions so they share a common denominator before we are able to add. We must do the same thing when adding or subtracting rational expressions.

The easiest common denominator to use will be the least common denominator    , or LCD. The LCD is the smallest multiple that the denominators have in common. To find the LCD of two rational expressions, we factor the expressions and multiply all of the distinct factors. For instance, if the factored denominators were $\text{\hspace{0.17em}}\left(x+3\right)\left(x+4\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(x+4\right)\left(x+5\right),$ then the LCD would be $\text{\hspace{0.17em}}\left(x+3\right)\left(x+4\right)\left(x+5\right).$

Once we find the LCD, we need to multiply each expression by the form of 1 that will change the denominator to the LCD. We would need to multiply the expression with a denominator of $\text{\hspace{0.17em}}\left(x+3\right)\left(x+4\right)\text{\hspace{0.17em}}$ by $\text{\hspace{0.17em}}\frac{x+5}{x+5}\text{\hspace{0.17em}}$ and the expression with a denominator of $\text{\hspace{0.17em}}\left(x+4\right)\left(x+5\right)\text{\hspace{0.17em}}$ by $\text{\hspace{0.17em}}\frac{x+3}{x+3}.$

Given two rational expressions, add or subtract them.

1. Factor the numerator and denominator.
2. Find the LCD of the expressions.
3. Multiply the expressions by a form of 1 that changes the denominators to the LCD.
4. Add or subtract the numerators.
5. Simplify.

$\frac{5}{x}+\frac{6}{y}$

First, we have to find the LCD. In this case, the LCD will be $\text{\hspace{0.17em}}xy.\text{\hspace{0.17em}}$ We then multiply each expression by the appropriate form of 1 to obtain $\text{\hspace{0.17em}}xy\text{\hspace{0.17em}}$ as the denominator for each fraction.

$\begin{array}{l}\frac{5}{x}\cdot \frac{y}{y}+\frac{6}{y}\cdot \frac{x}{x}\\ \frac{5y}{xy}+\frac{6x}{xy}\end{array}$

Now that the expressions have the same denominator, we simply add the numerators to find the sum.

$\frac{6x+5y}{xy}$

## Subtracting rational expressions

Subtract the rational expressions:

$\frac{6}{{x}^{2}+4x+4}-\frac{2}{{x}^{2}-4}$

Do we have to use the LCD to add or subtract rational expressions?

No. Any common denominator will work, but it is easiest to use the LCD.

Subtract the rational expressions: $\text{\hspace{0.17em}}\frac{3}{x+5}-\frac{1}{x-3}.$

$\frac{2\left(x-7\right)}{\left(x+5\right)\left(x-3\right)}$

## Simplifying complex rational expressions

A complex rational expression is a rational expression that contains additional rational expressions in the numerator, the denominator, or both. We can simplify complex rational expressions by rewriting the numerator and denominator as single rational expressions and dividing. The complex rational expression $\text{\hspace{0.17em}}\frac{a}{\frac{1}{b}+c}\text{\hspace{0.17em}}$ can be simplified by rewriting the numerator as the fraction $\text{\hspace{0.17em}}\frac{a}{1}\text{\hspace{0.17em}}$ and combining the expressions in the denominator as $\text{\hspace{0.17em}}\frac{1+bc}{b}.\text{\hspace{0.17em}}$ We can then rewrite the expression as a multiplication problem using the reciprocal of the denominator. We get $\text{\hspace{0.17em}}\frac{a}{1}\cdot \frac{b}{1+bc},$ which is equal to $\text{\hspace{0.17em}}\frac{ab}{1+bc}.$

The sequence is {1,-1,1-1.....} has
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