1960 AHSME Problems/Problem 9

Problem

The fraction $\frac{a^2+b^2-c^2+2ab}{a^2+c^2-b^2+2ac}$ is (with suitable restrictions of the values of a, b, and c):

$\text{(A) irreducible}\qquad$

$\text{(B) reducible to negative 1}\qquad$

$\text{(C) reducible to a polynomial of three terms}\qquad$

$\text{(D) reducible to} \frac{a-b+c}{a+b-c}\qquad$

$\text{(E) reducible to} \frac{a+b-c}{a-b+c}$

Solution

Use the commutative property to get $\frac{a^2+2ab+b^2-c^2}{a^2+2ac+c^2-b^2}$ Factor perfect square trinomials to get $\frac{(a+b)^2-c^2}{(a+c)^2-b^2}$ Factor difference of squares to get $\frac{(a+b+c)(a+b-c)}{(a+b+c)(a-b+c)}$ Cancel out like terms (with suitable restrictions of a, b, c) to get $\frac{a+b-c}{a-b+c}$ The answer is $\boxed{\textbf{(E)}}$ .

1960 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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1960 AHSME Problems/Problem 9

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