AHSME 1969 Problem 1

When $x$ is added to both the numerator and denominator of the fraction $\frac{a}{b},a \ne b,b \ne 0$, the value of the fraction is changed to $\frac{c}{d}$. Then $x$ equals: $\text{(A) } \frac{1}{c-d} \text{(B) } \frac{ad-bc}{c-d} \text{(C) } \frac{ad-bc}{c+d} \text{(D) }\frac{bc-ad}{c-d} \text{(E) } \frac{bc+ad}{c-d}$