1950 AHSME Problems/Problem 4

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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 start off by factoring the second fraction.

\[-\frac{ab-b^2}{ab-a^2} = -\frac{b(a-b)}{-a(a-b)} = \frac{b}{a}.\]

Now create a common denominator and simplify.

\[\frac{a^2-b^2}{ab}+\frac{b}{a}=\frac{a^2-b^2}{ab}+\frac{b^2}{ab} = \frac{a^2}{ab} = \boxed{\mathrm{(A) }\frac{a}{b}}\]

obs: Assume that $a \not = 0, b\not = 0, a\not = b$.

See Also

1950 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
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