# Difference between revisions of "1951 AHSME Problems/Problem 21"

## 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 $\textbf{(C)}\ xz > yz\implies x>y\text{ iff }x>0$, 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 $z^2$.

## See Also

 1951 AHSC (Problems • Answer Key • Resources) Preceded byProblem 20 Followed byProblem 22 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 All AHSME Problems and Solutions

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