Difference between revisions of "1951 AHSME Problems/Problem 21"
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== Solution == | == Solution == | ||
− | {{ | + | <math>\textbf{(A)}\ x \plus{} z > y \plus{} z\implies x>y</math>, just subtract <math>z</math> from both sides |
+ | |||
+ | <math>\textbf{(B)}\ x \minus{} z > y \minus{} z\implies x>y</math>, just add <math>z</math> to both sides | ||
+ | |||
+ | <math>\textbf{(C)}\ xz > yz\implies x>y\text{ iff }x>0</math>, so that means that our desired answer is <math>\boxed{\textbf{(C)}\ xz > yz}</math>. | ||
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+ | As a check: | ||
+ | |||
+ | <math>\textbf{(D)}\ \frac {x}{z^2} > \frac {y}{z^2}\implies x>y</math>, we can divide <math>z^2</math> safely and without worry because <math>z^2>0</math>. | ||
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+ | <math>\textbf{(E)}\ xz^2 > yz^2\implies x>y</math>, similar reasoning as above but instead, multiply <math>z^2</math>. | ||
== See Also == | == See Also == |
Revision as of 21:31, 10 April 2013
Problem
Given: and $z\not \equal{} 0$ (Error compiling LaTeX. Unknown error_msg). The inequality which is not always correct is:
$\textbf{(A)}\ x \plus{} z > y \plus{} z \qquad\textbf{(B)}\ x \minus{} z > y \minus{} z \qquad\textbf{(C)}\ xz > yz$ (Error compiling LaTeX. Unknown error_msg)
Solution
$\textbf{(A)}\ x \plus{} z > y \plus{} z\implies x>y$ (Error compiling LaTeX. Unknown error_msg), just subtract from both sides
$\textbf{(B)}\ x \minus{} z > y \minus{} z\implies x>y$ (Error compiling LaTeX. Unknown error_msg), just add to both sides
, so that means that our desired answer is
.
As a check:
, we can divide
safely and without worry because
.
, similar reasoning as above but instead, multiply
.
See Also
1951 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
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All AHSME Problems and Solutions |