1974 AHSME Problems/Problem 1

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Problem

If $x\not=0$ or $4$ and $y\not=0$ or $6$, then $\frac{2}{x}+\frac{3}{y}=\frac{1}{2}$ is equivalent to

$\mathrm{(A)\ } 4x+3y=xy \qquad \mathrm{(B) \ }y=\frac{4x}{6-y} \qquad \mathrm{(C) \  } \frac{x}{2}+\frac{y}{3}=2 \qquad$

$\mathrm{(D) \  } \frac{4y}{y-6}=x \qquad \mathrm{(E) \  }\text{none of these}$

Solution

To clear out the fractions, we multiply both sides of the equation by $xy$ to get $2y+3x=\frac{xy}{2}$, or $4y+6x=xy$. From this, we have $4y=xy-6x$. We can factor an $x$ out of the RHS, and so $4y=x(y-6)\implies \frac{4y}{y-6}=x, \boxed{\text{D}}$.

See Also

1974 AHSME (ProblemsAnswer KeyResources)
Preceded by
First Problem
Followed by
Problem 2
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