Home

Tutoring

Subjects

Live Classes

Study Coach

Essay Review

On-Demand Courses

Colleges

Games

Sign up

Log in

Opening subject page...

Loading your content

Practice

  • All Subjects
  • Algebra Flashcards
  • SAT Math Practice Tests
  • Math Question of the Day
  • Live Classes
  • On-Demand Courses

Varsity Tutors

  • Find a Tutor
  • Test Prep
  • Online Classes
  • K-12 Learning
  • College Search
  • VarsityTutors.com

© 2026 Varsity Tutors. All rights reserved.

← Back to quizzes

College Algebra Quiz

College Algebra Quiz: Complex Fractions

Practice Complex Fractions in College Algebra with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

Question 1 / 2

0 of 2 answered

What is the simplified form of 1a+1b1a−1b\frac{\frac{1}{a} + \frac{1}{b}}{\frac{1}{a} - \frac{1}{b}}a1​−b1​a1​+b1​​?

Select an answer to continue

What this quiz covers

This quiz focuses on Complex Fractions, giving you a quick way to practice the rules, question types, and explanations that matter most for College Algebra.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

What is the simplified form of 1a+1b1a−1b\frac{\frac{1}{a} + \frac{1}{b}}{\frac{1}{a} - \frac{1}{b}}a1​−b1​a1​+b1​​?

  1. a+ba−b\frac{a+b}{a-b}a−ba+b​
  2. b+ab−a\frac{b+a}{b-a}b−ab+a​ (correct answer)
  3. a+bb−a\frac{a+b}{b-a}b−aa+b​
  4. b−aa+b\frac{b-a}{a+b}a+bb−a​

Explanation: To simplify this complex fraction, first combine the fractions in the numerator and denominator. The numerator is 1a+1b=b+aab\frac{1}{a} + \frac{1}{b} = \frac{b+a}{ab}a1​+b1​=abb+a​. The denominator is 1a−1b=b−aab\frac{1}{a} - \frac{1}{b} = \frac{b-a}{ab}a1​−b1​=abb−a​. The complex fraction becomes b+aabb−aab=b+aab⋅abb−a=b+ab−a\frac{\frac{b+a}{ab}}{\frac{b-a}{ab}} = \frac{b+a}{ab} \cdot \frac{ab}{b-a} = \frac{b+a}{b-a}abb−a​abb+a​​=abb+a​⋅b−aab​=b−ab+a​. Choice A is incorrect because it has the wrong signs in both numerator and denominator. Choice C is incorrect because it has the wrong sign in the denominator. Choice D is incorrect because the numerator and denominator are swapped.

Question 2

Evaluate a−ba+b−a+ba−ba−ba+b+a+ba−b\frac{\frac{a-b}{a+b} - \frac{a+b}{a-b}}{\frac{a-b}{a+b} + \frac{a+b}{a-b}}a+ba−b​+a−ba+b​a+ba−b​−a−ba+b​​ and express in simplest form.

  1. −4ab2(a2+b2)\frac{-4ab}{2(a^2+b^2)}2(a2+b2)−4ab​
  2. −2aba2+b2\frac{-2ab}{a^2+b^2}a2+b2−2ab​ (correct answer)
  3. 4aba2−b2\frac{4ab}{a^2-b^2}a2−b24ab​
  4. −4aba2−b2\frac{-4ab}{a^2-b^2}a2−b2−4ab​

Explanation: First, find a common denominator for both the numerator and denominator of the complex fraction. For the numerator: a−ba+b−a+ba−b=(a−b)2−(a+b)2(a+b)(a−b)=a2−2ab+b2−(a2+2ab+b2)(a+b)(a−b)=−4aba2−b2\frac{a-b}{a+b} - \frac{a+b}{a-b} = \frac{(a-b)^2 - (a+b)^2}{(a+b)(a-b)} = \frac{a^2-2ab+b^2-(a^2+2ab+b^2)}{(a+b)(a-b)} = \frac{-4ab}{a^2-b^2}a+ba−b​−a−ba+b​=(a+b)(a−b)(a−b)2−(a+b)2​=(a+b)(a−b)a2−2ab+b2−(a2+2ab+b2)​=a2−b2−4ab​. For the denominator: a−ba+b+a+ba−b=(a−b)2+(a+b)2(a+b)(a−b)=a2−2ab+b2+a2+2ab+b2a2−b2=2a2+2b2a2−b2=2(a2+b2)a2−b2\frac{a-b}{a+b} + \frac{a+b}{a-b} = \frac{(a-b)^2 + (a+b)^2}{(a+b)(a-b)} = \frac{a^2-2ab+b^2+a^2+2ab+b^2}{a^2-b^2} = \frac{2a^2+2b^2}{a^2-b^2} = \frac{2(a^2+b^2)}{a^2-b^2}a+ba−b​+a−ba+b​=(a+b)(a−b)(a−b)2+(a+b)2​=a2−b2a2−2ab+b2+a2+2ab+b2​=a2−b22a2+2b2​=a2−b22(a2+b2)​. The complex fraction becomes: −4aba2−b22(a2+b2)a2−b2=−4aba2−b2⋅a2−b22(a2+b2)=−4ab2(a2+b2)=−2aba2+b2\frac{\frac{-4ab}{a^2-b^2}}{\frac{2(a^2+b^2)}{a^2-b^2}} = \frac{-4ab}{a^2-b^2} \cdot \frac{a^2-b^2}{2(a^2+b^2)} = \frac{-4ab}{2(a^2+b^2)} = \frac{-2ab}{a^2+b^2}a2−b22(a2+b2)​a2−b2−4ab​​=a2−b2−4ab​⋅2(a2+b2)a2−b2​=2(a2+b2)−4ab​=a2+b2−2ab​. Choice A is incorrect because it doesn't simplify the fraction. Choice C is incorrect because it has the wrong sign and wrong denominator. Choice D is incorrect because it has the wrong denominator.