If $\frac{x-y}{x+y}=5$, then $\frac{2 x+3 y}{3 x-2 y}$ equals
(A) 1
(B) 0
(C) $\frac{2}{3}$
(D) $\frac{15}{2}$
(E) $\frac{12}{5}$
Since $\frac{x-y}{x+y}=5$, then $x-y=5(x+y)$.

This means that $x-y=5 x+5 y$ and so $0=4 x+6 y$ or $2 x+3 y=0$.

Therefore, $\frac{2 x+3 y}{3 x-2 y}=\frac{0}{3 x-2 y}=0$.

(A specific example of $x$ and $y$ that works is $x=3$ and $y=-2$.

This gives $\frac{x-y}{x+y}=\frac{3-(-2)}{3+(-2)}=\frac{5}{1}=5$ and $\frac{2 x+3 y}{3 x-2 y}=\frac{2(3)+3(-2)}{3(3)-2(-2)}=\frac{0}{13}=0$.)

We should also note that if $\frac{x-y}{x+y}=5$, then $2 x+3 y=0$ (from above) and $x$ and $y$ cannot both be 0 (for $\frac{x-y}{x+y}$ to be well-defined).

Since $2 x+3 y=0$, then $x=-\frac{3}{2} y$ which means that it is not possible for only one of $x$ and $y$ to be 0 , which means that neither is 0 .

Also, the denominator of our desired expression $\frac{2 x+3 y}{3 x-2 y}$ is $\left.3 x-2 y=3\left(-\frac{3}{2} y\right)\right)-2 y=-\frac{13}{2} y$ which is not 0 , since $y \neq 0$.

Therefore, if $\frac{x-y}{x+y}=5$, then $\frac{2 x+3 y}{3 x-2 y}=0$.

ANSWER: (B)

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