# 1967 AHSME Problems/Problem 10

## Problem

If $\frac{a}{10^x-1}+\frac{b}{10^x+2}=\frac{2 \cdot 10^x+3}{(10^x-1)(10^x+2)}$ is an identity for positive rational values of $x$, then the value of $a-b$ is: $\textbf{(A)}\ \frac{4}{3} \qquad \textbf{(B)}\ \frac{5}{3} \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ \frac{11}{4} \qquad \textbf{(E)}\ 3$

## Solution

Given the equation: $$\frac{a}{10^x-1}+\frac{b}{10^x+2}=\frac{2 \cdot 10^x+3}{(10^x-1)(10^x+2)}$$

Let's simplify by letting $y = 10^x$. The equation becomes: $$\frac{a}{y-1}+\frac{b}{y+2}=\frac{2y+3}{(y-1)(y+2)}$$

Multiplying each term by the common denominator $(y-1)(y+2)$, we obtain: $$a(y+2) + b(y-1) = 2y+3$$

For the equation to be an identity, the coefficients of like terms on both sides must be equal. Equating the coefficients of $y$ and the constant terms, we get the system of equations: \begin{align*} a+b &= 2 \quad \text{(from coefficients of } y \text{)} \\ 2a - b &= 3 \quad \text{(from constant terms)} \end{align*}

Solving this system, we find: $$a = \frac{5}{3} \quad \text{and} \quad b = \frac{1}{3}$$

Thus, the difference is: $$a-b = \boxed{\textbf{(A) } \frac{4}{3}}$$

~ proloto

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 