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Complex Fractions - College Algebra

  1. What is the simplified form of 1a+1b1a−1b\frac{\frac{1}{a} + \frac{1}{b}}{\frac{1}{a} - \frac{1}{b}}?
    Answer: b+ab−a\frac{b+a}{b-a} 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}. The denominator is 1a−1b=b−aab\frac{1}{a} - \frac{1}{b} = \frac{b-a}{ab}. 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}. 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.
  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}} and express in simplest form.
    Answer: −2aba2+b2\frac{-2ab}{a^2+b^2} 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}. 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}. 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}. 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.

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