➗ Mathematics

Algebra 2: Rational Expressions and Equations

Simplify, add and solve rational expressions in Algebra 2, including domain restrictions and extraneous solutions, with worked examples.

  • ⏱ 3 min read
  • ✏️ 3 worked examples
  • 🎯 5 practice questions
  • 📶 Intermediate
High School

Curriculum fit: US Algebra 2. Aligned to Common Core HSA-APR.D.6 (rewrite rational expressions) and HSA-REI.A.2 (solve rational equations and check for extraneous solutions). Check your own state standards and course.

On this page

What Is It?

A rational expression is a fraction with polynomials on the top and bottom, such as (x + 3) ÷ (x + 4). Because you can never divide by zero, some values of x are not allowed.

To simplify, factor the top and bottom and cancel common factors. To add or subtract, find a common denominator first. To solve an equation, multiply both sides by the common denominator, solve, then check each answer.

Checking matters: a solution that makes a denominator zero is extraneous and must be thrown out.

FactordenominatorsFind LCDand multiplyboth sidesSolvethe resultingequationCheckreject if thedenominator is 0
Factor, find the common denominator, clear the fractions, solve, and then reject any extraneous solution.

Key Formulas and Ideas

  • Simplify: factor, then cancel common factors (not common terms).
  • Add: a/b + c/d = (ad + bc) ÷ bd, or use the lowest common denominator (LCD).
  • Multiply: multiply the tops and the bottoms, after factoring.
  • Divide: multiply by the reciprocal of the second fraction.
  • Domain: exclude any x that makes a denominator equal to 0.

Worked Examples

Example 1

Simplify and State the Restrictions

Question

Simplify (x² − 9) ÷ (x² + x − 12).

  1. Factor: (x − 3)(x + 3) ÷ [(x + 4)(x − 3)].
  2. Cancel the common factor x − 3, leaving (x + 3) ÷ (x + 4).
  3. The original denominator is zero when x = 3 or x = −4, so these are excluded.

✓ Answer (x + 3) ÷ (x + 4), with x ≠ 3 and x ≠ −4

Example 2

Add Two Rational Expressions

Question

Simplify 1/x + 1/(x + 2).

  1. The LCD is x(x + 2).
  2. Rewrite: (x + 2) ÷ [x(x + 2)] + x ÷ [x(x + 2)].
  3. Add the tops: (2x + 2) ÷ [x(x + 2)].

✓ Answer (2x + 2) ÷ [x(x + 2)]

Example 3

Solve and Check for Extraneous Solutions

Question

Solve x² ÷ (x − 3) = 9 ÷ (x − 3).

  1. Multiply both sides by x − 3 (x ≠ 3): x² = 9.
  2. So x = 3 or x = −3.
  3. x = 3 makes the denominator zero, so reject it.

✓ Answer x = −3

Common Mistakes to Avoid

  • Cancelling terms instead of factors. In (x + 3) ÷ (x + 4) you cannot cancel the x.
  • Forgetting to state the values that x cannot take.
  • Not checking answers, and keeping an extraneous solution.
  • Adding fractions by adding the denominators.

Practice Questions

0 of 5 answered
Q1What value is excluded from the domain of 1 ÷ (x − 5)?
Q2Simplify 2x ÷ x² (x ≠ 0).
Q3What is the LCD of 1/x and 1/(x + 1)?
Q4Solve 3 ÷ (x − 1) = 2 ÷ (x + 1).
Q5An extraneous solution is one that…

Frequently Asked Questions

What is an extraneous solution?

It is an answer that comes out of your algebra but does not work in the original equation, usually because it makes a denominator zero. Always substitute back to check.

Can I cancel the x in (x + 3) ÷ (x + 4)?

No. You can only cancel factors that multiply the whole top and the whole bottom. The x here is part of a sum, so it cannot be cancelled.

How do I find the LCD?

Factor each denominator, then take every factor the greatest number of times it appears in any denominator.

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Sources and Review

Reviewed by StudyMorrow editorial review on 6 Nov 2026.