1960 AHSME Problems/Problem 14

Problem

If $a$ and $b$ are real numbers, the equation $3x-5+a=bx+1$ has a unique solution $x$ [The symbol $a \neq 0$ means that $a$ is different from zero]:

$\text{(A) for all a and b} \qquad \text{(B) if a }\neq\text{2b}\qquad \text{(C) if a }\neq 6\qquad \\ \text{(D) if b }\neq 0\qquad \text{(E) if b }\neq 3$

Solution

If the coefficients of the x-terms are equal on both sides, then when the x-terms are subtracted from both sides, the equation results in a number equals 1.

This means the equation has either infinite or no solutions, so the x-terms can not be equal on both sides. Thus, $b \neq 3$ , so the answer is $\boxed{\textbf{(E)}}$ .

1960 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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1960 AHSME Problems/Problem 14

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