Track / Stage 5 / 125 of 400 #725 of 1964

Problem 725

AIME late

Algebra Difficulty 5.2 Find the answer

Calculate the value of the expression $\frac{2 a-b}{3 a-b}+\frac{5 b-a}{3 a+b}$ , given that $10 a^{2}-3 b^{2}+5 a b=0$ and $9 a^{2}-b^{2} \neq 0$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Official solution

Solution. Since $5 a b=3 b^{2}-10 a^{2}$ , then

$\begin{aligned} & \quad \frac{2 a-b}{3 a-b}+\frac{5 b-a}{3 a+b}=\frac{3 a^{2}+15 a b-6 b^{2}}{9 a^{2}-b^{2}}=\frac{3 a^{2}+3\left(3 b^{2}-10 a^{2}\right)-6 b^{2}}{9 a^{2}-b^{2}}= \\ & \text { Answer: }-3 . \\ & =\frac{-3\left(9 a^{2}-b^{2}\right)}{9 a^{2}-b^{2}}=-3 . \end{aligned}$

Source: NuminaMath-1.5, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.