# 1951 AHSME Problems/Problem 21

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## Problem

Given: $x > 0, y > 0, x > y$ and $z\ne 0$. The inequality which is not always correct is: $\textbf{(A)}\ x + z > y + z \qquad\textbf{(B)}\ x - z > y - z \qquad\textbf{(C)}\ xz > yz$ $\textbf{(D)}\ \frac {x}{z^2} > \frac {y}{z^2} \qquad\textbf{(E)}\ xz^2 > yz^2$

## Solution $\textbf{(A)}\ x + z > y + z\implies x>y$, just subtract $z$ from both sides $\textbf{(B)}\ x - z > y - z\implies x>y$, just add $z$ to both sides $\textbf{(C)}\ xz > yz\implies x>y\text{ if }x>0$, so that means that our desired answer is $\boxed{\textbf{(C)}\ xz > yz}$.

As a check: $\textbf{(D)}\ \frac {x}{z^2} > \frac {y}{z^2}\implies x>y$, we can divide $z^2$ safely and without worry because $z^2>0$. $\textbf{(E)}\ xz^2 > yz^2\implies x>y$, similar reasoning as above but instead, multiply $z^2$.

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