Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 1.6 Junior Find the answer

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

Pick one

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 $4y=x(y-6)\implies \frac{4y}{y-6}=x, \boxed{\text{D}}$ .

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