# Difference between revisions of "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 $\boxed{\textbf{(E)}}$.

## See Also

 1960 AHSC (Problems • Answer Key • Resources) Preceded byProblem 13 Followed byProblem 15 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 All AHSME Problems and Solutions
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