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

1967 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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1967 AHSME Problems/Problem 10

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