# Rational expressions and equations

Canonical: https://duckyhelper.com/learn/algebra-2/rational-expressions/
Updated: 2026-10-01

A rational expression is a fraction with polynomials on the top and bottom, like \(\frac{x^2 - 9}{x^2 + x - 6}\). Simplify it by factoring both and canceling common factors, never single terms. Multiply straight across, divide by flipping the second fraction, and add or subtract over a common denominator. To solve a rational equation, multiply by the LCD, solve, then throw out any answer that makes a denominator zero.

## 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.

1. Factor the top (difference of squares) and the bottom (trinomial).

   $$
   \frac{x^2 - 9}{x^2 + x - 6} = \frac{(x - 3)(x + 3)}{(x + 3)(x - 2)}
   $$
2. Cancel the common factor \(x + 3\).

   $$
   \frac{(x - 3)(x + 3)}{(x + 3)(x - 2)} = \frac{x - 3}{x - 2}
   $$
3. 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 \(\frac{2}{x^2 - 1} - \frac{1}{x - 1}\).

1. Factor the first denominator: \(x^2 - 1 = (x - 1)(x + 1)\). The LCD is \((x - 1)(x + 1)\). Multiply the second fraction by \(\frac{x + 1}{x + 1}\).

   $$
   \frac{2}{x^2 - 1} - \frac{1}{x - 1} = \frac{2}{(x - 1)(x + 1)} - \frac{x + 1}{(x - 1)(x + 1)}
   $$
2. Subtract the whole second numerator. Use parentheses.

   $$
   \frac{2}{(x - 1)(x + 1)} - \frac{x + 1}{(x - 1)(x + 1)} = \frac{2 - (x + 1)}{(x - 1)(x + 1)}
   $$
3. Simplify the top: \(2 - x - 1 = 1 - x\).

   $$
   \frac{2 - (x + 1)}{(x - 1)(x + 1)} = \frac{1 - x}{(x - 1)(x + 1)}
   $$
4. \(1 - x\) 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 \(\frac{x^2 - 4}{x^2 + 3x} \div \frac{x + 2}{x^2 - 9}\).

1. Flip the second fraction and multiply.

   $$
   \frac{x^2 - 4}{x^2 + 3x} \div \frac{x + 2}{x^2 - 9} = \frac{x^2 - 4}{x^2 + 3x} \cdot \frac{x^2 - 9}{x + 2}
   $$
2. Factor every piece.

   $$
   \frac{x^2 - 4}{x^2 + 3x} \cdot \frac{x^2 - 9}{x + 2} = \frac{(x - 2)(x + 2)}{x(x + 3)} \cdot \frac{(x - 3)(x + 3)}{x + 2}
   $$
3. 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 \(\frac{x}{x - 3} - \frac{1}{x} = \frac{3}{x(x - 3)}\).

1. Restrictions first: \(x \ne 0\) and \(x \ne 3\). The LCD is \(x(x - 3)\). Multiply every term by it.

   $$
   x^2 - (x - 3) = 3
   $$
2. Simplify.

   $$
   x^2 - x + 3 = 3
   $$
3. Subtract 3 and factor.

   $$
   x(x - 1) = 0
   $$
4. 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

> **Tip: Spot the opposite factor**
>
> \(a - b = -(b - a)\). So \(\frac{1 - x}{x - 1} = -1\). When two factors look almost the same but reversed, pull out \(-1\) and they cancel.

> **Note: Check by plugging in a number**
>
> Pick a value that is not restricted, like \(x = 5\), and put it into the original and your answer. If the two values differ, there is a mistake. If they match, you are very likely right. This is a check, not a proof.

## Practice

**5 practice questions**

1. Simplify \(\frac{x^2 - 5x}{x^2 - 25}\).
   A. \(\frac{x}{x + 5}\)
   B. \(\frac{x}{5}\)
   C. \(\frac{1}{x + 5}\)
   D. \(\frac{x}{x - 5}\)

   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.

2. Which is equal to \(\frac{3}{x} + \frac{2}{x + 4}\)?
   A. \(\frac{5}{2x + 4}\)
   B. \(\frac{5x + 12}{x(x + 4)}\)
   C. \(\frac{5x + 4}{x(x + 4)}\)
   D. \(\frac{5}{x(x + 4)}\)

   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.

3. For which values of x is \(\frac{x + 2}{x^2 - 2x - 8}\) undefined?
   A. \(4\) and \(-2\)
   B. \(4\) only
   C. \(-2\) only
   D. \(-4\) and \(2\)

   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 \(-2\).

4. Solve \(\frac{4}{x - 1} = \frac{x + 3}{x - 1}\).
   A. \(x = 1\)
   B. \(x = 4\)
   C. \(x = -1\)
   D. No solution

   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.

5. Solve \(\frac{2}{x + 1} + \frac{1}{x - 1} = \frac{4}{x^2 - 1}\). Enter your answer as a fraction.

   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 \((x - 1)(x + 1)\).

## Related

- [How to factor polynomials](https://duckyhelper.com/learn/algebra-1/factoring/)
- [Dividing polynomials](https://duckyhelper.com/learn/algebra-2/polynomial-division/)
- [Nonlinear equations and systems on the SAT](https://duckyhelper.com/learn/sat-math/nonlinear-equations-and-systems/)
- [Equivalent expressions on the SAT](https://duckyhelper.com/learn/sat-math/equivalent-expressions/)
- [Algebra 2 study guides](https://duckyhelper.com/learn/algebra-2/)

## Try asking Ducky

- "Why is x = 0 not allowed when the algebra says it works?"
- "I keep getting 3 + x on top when I subtract. Where is my sign wrong?"
- "Give me three rational equations, one of them with no solution."

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