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College Algebra

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.

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.

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​
  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​
  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.