# 1977 AHSME Problems/Problem 6

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## Problem 6

If $x, y$ and $2x + \frac{y}{2}$ are not zero, then $\left( 2x + \frac{y}{2} \right)^{-1} \left[(2x)^{-1} + \left( \frac{y}{2} \right)^{-1} \right]$ equals $\textbf{(A) }1\qquad \textbf{(B) }xy^{-1}\qquad \textbf{(C) }x^{-1}y\qquad \textbf{(D) }(xy)^{-1}\qquad \textbf{(E) }\text{none of these}$

## Solution

We can write $\left( 2x+ \frac{y}{2} \right)^{-1} = \left( \frac{4x+y}{2} \right)^{-1} = \frac{2}{4x+y}$.

Then, the expression simplifies to $$\left( 2x + \frac{y}{2} \right)^{-1} \left[(2x)^{-1} + \left( \frac{y}{2} \right)^{-1} \right] \Rightarrow \frac{2}{4x+y} \left( \frac{1}{2x} + \frac{2}{y} \right) \Rightarrow \frac{2}{4x+y} \left( \frac{y+4x}{2xy} \right) \Rightarrow \frac{1}{xy}.$$

Thus, our answer is $\textbf{(D) }(xy)^{-1}$. ~jiang147369

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