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Difference between revisions of "1985 AHSME Problems/Problem 8"

(Created page with "==Problem== Let <math> a, a', b, </math> and <math> b' </math> be real numbers with <math> a </math> and <math> a' </math> nonzero. The solution to <math> ax+b=0 </math> is less ...")
(No difference)

Revision as of 20:37, 7 October 2011

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 (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
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
All AHSME Problems and Solutions
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