If $\frac{a}{b}=3$ and $\frac{b}{c}=2$, then the value of $\frac{a-b}{c-b}$ is
(A) -4
(B) $-\frac{1}{3}$
(C) $\frac{2}{3}$
(D) 2
(E) 6
Solution 1

Since $\frac{a}{b}=3$, then $a=3 b$. Since $\frac{b}{c}=2$, then $b=2 c$.

Since $a=3 b$ and $b=2 c$, then $a \stackrel{c}{=} 6 c$.

Therefore, $\frac{a-b}{c-b}=\frac{6 c-2 c}{c-2 c}=\frac{4 c}{-c}=-4$.

## Solution 2

We divide the numerator and denominator of the given expression by $b$ to obtain $\frac{a-b}{c-b}=\frac{\frac{a}{b}-1}{\frac{c}{b}-1}$.

Since $\frac{b}{c}=2$, then $\frac{c}{b}=\frac{1}{2}$.

Therefore, $\frac{a-b}{c-b}=\frac{3-1}{\frac{1}{2}-1}=\frac{2}{-\frac{1}{2}}=-4$.

## Solution 3

Try $c=1$. Since $\frac{b}{c}=2$, then $b=2$. Since $\frac{a}{b}=3$, then $a=6$.

Therefore, $\frac{a-b}{c-b}=\frac{6-2}{1-2}=\frac{4}{-1}=-4$.

ANSWER: (A)