Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 4.6 AIME Prove it Slovenia

For real numbers $a$ and $b$ , such that $|a| \neq |b|$ and $a \neq 0$ , we have
$\frac{a-b}{a^2+ab} + \frac{a+b}{a^2-ab} = \frac{3a-b}{a^2-b^2}.$
Determine the value of the expression $\frac{b}{a}$ .

Solution

Multiplying the equation by $a(a+b)(a-b)$ we get
$(a-b)^2 + (a+b)^2 = a(3a-b).$
After expanding the terms and moving all the terms to the right-hand side we get
$0 = a^2 - ab - 2b^2 = (a - 2b)(a + b).$
Since $a \neq -b$ , we get $a - 2b = 0$ or $a = 2b$ . Since $a$ is non-zero, we get $\frac{b}{a} = \frac{1}{2}$ .

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