1950 AHSME Problems/Problem 4

Problem

Reduced to lowest terms, $\frac{a^{2}-b^{2}}{ab}$ - $\frac{ab-b^{2}}{ab-a^{2}}$ is equal to:

$\textbf{(A)}\ \frac{a}{b}\qquad\textbf{(B)}\ \frac{a^{2}-2b^{2}}{ab}\qquad\textbf{(C)}\ a^{2}\qquad\textbf{(D)}\ a-2b\qquad\textbf{(E)}\ \text{None of these}$

Solution

We can use difference of two squares to expand $a^2-b^2=(a-b)(a+b)$ and factor to get $ab-b^{2}=b(a-b)$ and $a^2-ab=a(a-b)$

$\dfrac{a^{2}-b^{2}}{ab} - \dfrac{ab-b^{2}}{ab-a^{2}} =\dfrac{(a-b)(a+b)}{ab}-\dfrac{b(a-b)}{a(a-b)}$

We can further factor to get $\dfrac{(a-b)(a+b)}{ab}-\dfrac{b(a-b)}{a(a-b)}$ If we assume b is not equal to a, this is equal to $\dfrac{(a-b)(a+b)}{ab}-\dfrac{b}{a}=\dfrac{(a-b)(a+b)}{ab}-\dfrac{b^2}{ab}=\dfrac{(a-b)(a+b)-b^2}{ab}=\dfrac{a^2-2b^2}{ab}$

The answer is $\boxed{\textbf{(B)}\ \frac{a^{2}-2b^{2}}{ab}}$

1950 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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1950 AHSME Problems/Problem 4

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