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
Answer Choices:
A. a′b<ab′
B. ab′<a′b
C. ab<a′b′
D. ab<a′b′
E. a′b′<ab
Solution:
The solution of ax+b=0 is −b/a. The solution of a′x+b′=0 is −b′/a′. Thus the question becomes: which one of the five inequalities in the answers is equivalent to
(*) a−b<a′−b′?
Multiplying by −1, one obtains
ab>a′b′
which is (E). Conversely, multiplying (E) by −1 gives (∗), so they are equivalent.
Note. Multiplying (∗) by −aa′ results in (B) if, but only if, aa′>0. Thus (∗) and (B) are not equivalent.