When x is added to both the numerator and the denominator of the fraction (a)/(b), a≠ b, b≠0, the value of the fraction is changed to (c)/(d). Then x equals:

1969 AMC 12 Problem 1

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When

x

is added to both the numerator and the denominator of the fraction

\frac{a}{b},

a\ne b,

b\ne0,

the value of the fraction is changed to

\frac{c}{d}.

Then

x

equals:

\dfrac1{c-d}

\dfrac{ad-bc}{c-d}

\dfrac{ad-bc}{c+d}

\dfrac{bc-ad}{c-d}

\dfrac{bc-ad}{c+d}

Translate the change into

\frac{a+x}{b+x}=\frac{c}{d}

Cross-multiply and collect the terms containing

x

From

\frac{a+x}{b+x}=\frac cd

we obtain

ad+dx=bc+cx.

Hence

x(d-c)=bc-ad,

x=\frac{bc-ad}{d-c}=\frac{ad-bc}{c-d}.

Therefore, the correct answer is B.