8. Let $a$, $a^{\prime}$, $b$, $b^{\prime}$ be real numbers, and $a$ and $a^{\prime}$ be non-zero real numbers. Then the solution to the equation $a x + b = 0$ is less than the solution to the equation $a^{\prime} x + b^{\prime} = 0$ if and only if
(A) $a^{\prime} b < a b^{\prime}$;
(B) $a b^{\prime} < a^{\prime} b$,
(C) $a b < a^{\prime} b^{\prime}$;
(D) $\frac{b}{a} < \frac{b^{\prime}}{a^{\prime}}$;
(E) $\frac{b^{\prime}}{a^{\prime}} < \frac{b}{a}$.
8. The solution to the equation $a x+b=0$ is $-\frac{b}{a}$, and the solution to the equation $a^{\prime} x+b^{\prime}=0$ is $-\frac{b^{\prime}}{a^{\prime}}$. Thus, the problem becomes: among the five given answers, which one is equivalent to the inequality $-\frac{b}{a}a^{\prime} b^{\prime}$, which is $(E)$. Conversely, multiplying $(E)$ by -1 on both sides yields $-\frac{b}{a}<-\frac{b^{\prime}}{a^{\prime}} \cdot \|_{1},-\frac{b}{a}$ $<-\frac{b^{\prime}}{a^{\prime}}$ is equivalent to $_{a^{\prime}}^{b^{\prime}}<\frac{b}{a}$.
(E) is correct.