1985 AHSME Problems/Problem 8

Problem

Let $a,a',b,b'$ be real numbers with $a$ and $a'$ nonzero. The solution to $ax+b=0$ is less than the solution to $a'x+b'=0$ if and only if

$\mathrm{(A)\ } a'b < ab' \qquad \mathrm{(B) \ }ab' < a'b \qquad \mathrm{(C) \ } ab < a'b' \qquad \mathrm{(D) \ } \frac{b}{a} < \frac{b'}{a'} \qquad \mathrm{(E) \ }\frac{b'}{a'} < \frac{b}{a}$

Solution

The solution to $ax+b=0$ is $x = \frac{-b}{a}$ , while that to $a'x+b'=0$ is $x = \frac{-b'}{a'}$ . The first solution is less than the second precisely if $\frac{-b}{a} < \frac{-b'}{a'},$ and multiplying this inequality by $-1$ reversees the inequality sign, yielding $\boxed{\text{(E)} \ \frac{b'}{a'} < \frac{b}{a}}$ .

1985 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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1985 AHSME Problems/Problem 8

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