$2 \cdot 116$ If $x$ is added to both the numerator and denominator of the fraction $\frac{a}{b}(a \neq b, b \neq 0)$, the fraction becomes $\frac{c}{d}$, then $x$ equals
(A) $\frac{1}{c-d}$.
(B) $\frac{a d-b c}{c-d}$.
(C) $\frac{a d-b c}{c+d}$.
(D) $\frac{b c-a d}{c-d}$.
(E) $\frac{b c-a d}{c+d}$.
(20th American High School Mathematics Examination, 1969)
[Solution] According to the problem, we have $\frac{a+x}{b+x}=\frac{c}{d}$, then $a d+d x=b c+c x$, thus
$$
x=\frac{a d-b c}{c-d} \text {. }
$$

Therefore, the answer is $(B)$.