Rational expressions: simplify, add, solve

The key idea

Rational expressions follow the same rules as number fractions. The new part is factoring: you can only cancel something that multiplies the whole top and the whole bottom.

\frac{(x - 3)(x + 3)}{(x + 3)(x - 2)} = \frac{x - 3}{x - 2}, \quad x \ne -3,\ 2

The restrictions matter. The original expression is undefined at every value that makes its denominator zero, even if that factor cancels later.

The four operations
Operation	Rule	Then
Multiply	$\frac{A}{B} \cdot \frac{C}{D} = \frac{AC}{BD}$	Factor everything and cancel
Divide	$\frac{A}{B} \div \frac{C}{D} = \frac{A}{B} \cdot \frac{D}{C}$	Flip the second fraction, then multiply
Add or subtract	$\frac{A}{D} \pm \frac{C}{D} = \frac{A \pm C}{D}$	First rewrite both over the LCD
Solve an equation	Multiply every term by the LCD	Check answers against the restrictions

Worked examples

Example 1: simplify

Problem Simplify $\frac{x^2 - 9}{x^2 + x - 6}$ and state the restrictions.

Factor the top (difference of squares) and the bottom (trinomial).
$\frac{x^2 - 9}{x^2 + x - 6} = \frac{(x - 3)(x + 3)}{(x + 3)(x - 2)}$
Cancel the common factor $x + 3$ .
$\frac{(x - 3)(x + 3)}{(x + 3)(x - 2)} = \frac{x - 3}{x - 2}$
Restrictions come from the original bottom: $(x + 3)(x - 2) = 0$ at $x = -3$ and $x = 2$ .

Answer $\frac{x - 3}{x - 2}$ , with $x \ne -3$ and $x \ne 2$

Example 2: subtract with a common denominator

Problem Simplify $\frac{2}{x^2 - 1} - \frac{1}{x - 1}$ .

Factor the first denominator: $x^2 - 1 = (x - 1)(x + 1)$ . The LCD is $(x - 1)(x + 1)$ . Multiply the second fraction by $\frac{x + 1}{x + 1}$ .
$\frac{2}{x^2 - 1} - \frac{1}{x - 1} = \frac{2}{(x - 1)(x + 1)} - \frac{x + 1}{(x - 1)(x + 1)}$
Subtract the whole second numerator. Use parentheses.
$\frac{2}{(x - 1)(x + 1)} - \frac{x + 1}{(x - 1)(x + 1)} = \frac{2 - (x + 1)}{(x - 1)(x + 1)}$
Simplify the top: $2 - x - 1 = 1 - x$ .
$\frac{2 - (x + 1)}{(x - 1)(x + 1)} = \frac{1 - x}{(x - 1)(x + 1)}$
$1 - x$ is $-(x - 1)$ , so it cancels with $x - 1$ and leaves a minus sign.
$\frac{1 - x}{(x - 1)(x + 1)} = -\frac{1}{x + 1}$

Answer $-\frac{1}{x + 1}$ , with $x \ne 1$ and $x \ne -1$

Example 3: divide

Problem Simplify $\frac{x^2 - 4}{x^2 + 3x} \div \frac{x + 2}{x^2 - 9}$ .

Flip the second fraction and multiply.
$\frac{x^2 - 4}{x^2 + 3x} \div \frac{x + 2}{x^2 - 9} = \frac{x^2 - 4}{x^2 + 3x} \cdot \frac{x^2 - 9}{x + 2}$
Factor every piece.
$\frac{x^2 - 4}{x^2 + 3x} \cdot \frac{x^2 - 9}{x + 2} = \frac{(x - 2)(x + 2)}{x(x + 3)} \cdot \frac{(x - 3)(x + 3)}{x + 2}$
Cancel $x + 2$ and $x + 3$ .
$\frac{(x - 2)(x + 2)}{x(x + 3)} \cdot \frac{(x - 3)(x + 3)}{x + 2} = \frac{(x - 2)(x - 3)}{x}$

Answer $\frac{(x - 2)(x - 3)}{x}$

Example 4 (test-hard): an equation with an extraneous solution

Problem Solve $\frac{x}{x - 3} - \frac{1}{x} = \frac{3}{x(x - 3)}$ .

Restrictions first: $x \ne 0$ and $x \ne 3$ . The LCD is $x(x - 3)$ . Multiply every term by it.
$x^2 - (x - 3) = 3$
Simplify.
$x^2 - x + 3 = 3$
Subtract 3 and factor.
$x(x - 1) = 0$
So $x = 0$ or $x = 1$ . But $x = 0$ makes a denominator zero, so it is extraneous. Check $x = 1$ : the left side is $\frac{1}{-2} - 1 = -\frac{3}{2}$ , and the right side is $\frac{3}{-2} = -\frac{3}{2}$ .

Answer $x = 1$ ( $x = 0$ is extraneous)

Common mistakes

Canceling terms instead of factors. In $\frac{x + 6}{x + 2}$ , you cannot cancel the x's. Fix: only cancel a factor that multiplies the entire top and entire bottom.
Forgetting to distribute the minus when subtracting. $2 - (x + 1)$ is $1 - x$ , not $3 + x$ or $1 + x$ . Fix: put the second numerator in parentheses.
Dropping restrictions after canceling. $\frac{x - 3}{x - 2}$ came from an expression undefined at $x = -3$ . Fix: list restrictions from the original denominators before you cancel.
Keeping extraneous solutions. Multiplying by the LCD can create answers that make a denominator zero. Fix: compare every answer to your restriction list.
Flipping the wrong fraction when dividing. Only the second fraction (the divisor) flips. Fix: rewrite $\div$ as $\cdot$ and flip what comes after it.

Quick methods

Practice

5 practice questions

Simplify $\frac{x^2 - 5x}{x^2 - 25}$ .
1. $\frac{x}{x + 5}$
2. $\frac{x}{5}$
3. $\frac{1}{x + 5}$
4. $\frac{x}{x - 5}$
Show answer
Answer: $\frac{x}{x + 5}$
Factor: $\frac{x(x - 5)}{(x - 5)(x + 5)} = \frac{x}{x + 5}$ . $\frac{x}{5}$ comes from canceling the $x^2$ terms, which are not factors.
Which is equal to $\frac{3}{x} + \frac{2}{x + 4}$ ?
1. $\frac{5}{2x + 4}$
2. $\frac{5x + 12}{x(x + 4)}$
3. $\frac{5x + 4}{x(x + 4)}$
4. $\frac{5}{x(x + 4)}$
Show answer
Answer: $\frac{5x + 12}{x(x + 4)}$
Over the LCD $x(x + 4)$ : $\frac{3(x + 4) + 2x}{x(x + 4)} = \frac{5x + 12}{x(x + 4)}$ . $\frac{5}{2x + 4}$ adds tops and bottoms, which never works for fractions.
For which values of x is $\frac{x + 2}{x^2 - 2x - 8}$ undefined?
1. $4$ and $-2$
2. $4$ only
3. $-2$ only
4. $-4$ and $2$
Show answer
Answer: $4$ and $-2$
The bottom is $(x - 4)(x + 2)$ , which is 0 at $x = 4$ and $x = -2$ . The $x + 2$ cancels when you simplify, but the original expression is still undefined at $-2$ .
Solve $\frac{4}{x - 1} = \frac{x + 3}{x - 1}$ .
1. $x = 1$
2. $x = 4$
3. $x = -1$
4. No solution
Show answer
Answer: No solution
Multiplying by $x - 1$ gives $4 = x + 3$ , so $x = 1$ . But $x = 1$ makes both denominators zero, so it is extraneous and there is no solution.
Solve $\frac{2}{x + 1} + \frac{1}{x - 1} = \frac{4}{x^2 - 1}$ . Enter your answer as a fraction.
Show answer
Answer: 5/3
Multiply by $(x + 1)(x - 1)$ : $2(x - 1) + (x + 1) = 4$ , so $3x - 1 = 4$ and $x = \frac{5}{3}$ . It is not 1 or $-1$ , so it is allowed.

Frequently asked questions

Why can't I cancel terms in a fraction?

Canceling is dividing the top and bottom by the same thing. That only works when the thing multiplies everything on top and everything on bottom. In $\frac{x + 6}{x + 2}$ , x is added, not multiplied, so you would be changing the value. Try $x = 2$ : $\frac{8}{4} = 2$ , but canceling gives $\frac{6}{2} = 3$ .

What is an extraneous solution?

It is an answer you get from correct algebra that does not work in the original equation. With rational equations it happens when an answer makes a denominator zero. Multiplying both sides by an expression that can be zero is what lets it sneak in.

How do I find the LCD of rational expressions?

Factor every denominator. The LCD uses each different factor the greatest number of times it appears in any one denominator. For $x^2 - 1$ and $x - 1$ , the factors are $x - 1$ and $x + 1$ , so the LCD is $(x - 1)(x + 1)$ .

Rational expressions and equations

The key idea

Worked examples

Common mistakes

Quick methods

Practice

Frequently asked questions

Try asking Ducky

The key idea

Worked examples

Common mistakes

Quick methods

Practice

Frequently asked questions

Related

Try asking Ducky