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 $\boxed{\text{(E)} \ \frac{b'}{a'} < \frac{b}{a}}$.

See Also

1985 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
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