1985 AHSME Problems/Problem 8

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Problem

Let $a, a', b,$ and $b'$ 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 $\frac{-b}{a}$ and the solution to $a'x+b'=0$ is $\frac{-b'}{a'}$. The first solution is less than the second if $\frac{-b}{a}<\frac{-b'}{a'}$. We can get rid of the negative signs if we reverse the inequality sign, so we have $\frac{b'}{a'}<\frac{b}{a}$. All our steps are reversible, so our answer is $\boxed{\text{E}}$.

See Also

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