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1964 AHSME Problems/Problem 4 - Revision history
2024-03-29T12:58:54Z
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Talkinaway: Created page with "== Problem == The expression <cmath>\frac{P+Q}{P-Q}-\frac{P-Q}{P+Q}</cmath> where <math>P=x+y</math> and <math>Q=x-y</math>, is equivalent to: <math>\textbf{(A)}\ \frac{x^..."
2019-07-23T07:49:31Z
<p>Created page with "== Problem == The expression <cmath>\frac{P+Q}{P-Q}-\frac{P-Q}{P+Q}</cmath> where <math>P=x+y</math> and <math>Q=x-y</math>, is equivalent to: <math>\textbf{(A)}\ \frac{x^..."</p>
<p><b>New page</b></p><div>== Problem ==<br />
<br />
The expression<br />
<br />
<cmath>\frac{P+Q}{P-Q}-\frac{P-Q}{P+Q}</cmath><br />
<br />
where <math>P=x+y</math> and <math>Q=x-y</math>, is equivalent to:<br />
<br />
<math>\textbf{(A)}\ \frac{x^2-y^2}{xy}\qquad<br />
\textbf{(B)}\ \frac{x^2-y^2}{2xy}\qquad<br />
\textbf{(C)}\ 1 \qquad<br />
\textbf{(D)}\ \frac{x^2+y^2}{xy}\qquad<br />
\textbf{(E)}\ \frac{x^2+y^2}{2xy} </math> <br />
<br />
== Solution 1 ==<br />
<br />
Simplifying before substituting by obtaining a common denominator gives: <br />
<br />
<math>\frac{(P+Q)^2 - (P - Q)^2}{(P-Q)(P+Q)}</math><br />
<br />
<math>\frac{(P^2 + 2PQ + Q^2) - (P^2 - 2PQ + Q^2)}{P^2 - Q^2}</math><br />
<br />
<math>\frac{4PQ}{P^2 - Q^2}</math><br />
<br />
Substituting gives:<br />
<br />
<math>\frac{4(x+y)(x-y)}{(x+y)^2 - (x - y)^2}</math><br />
<br />
<math>\frac{4(x^2 - y^2)}{(x^2 + 2xy + y^2) - (x^2 - 2xy + y^2)}</math><br />
<br />
<math>\frac{4(x^2 - y^2)}{4xy}</math><br />
<br />
<math>\frac{x^2 - y^2}{xy}</math><br />
<br />
<math>\boxed{\textbf{(A)}}</math><br />
<br />
== Solution 2 ==<br />
<br />
Substituting and then simplifying, noting that <math>P+Q = 2x</math> and <math>P - Q = 2y</math>, gives:<br />
<br />
<math>\frac{2x}{2y} - \frac{2y}{2x}</math><br />
<br />
<math>\frac{x^2}{xy} - \frac{y^2}{xy}</math><br />
<br />
<math>\frac{x^2 - y^2}{xy}</math><br />
<br />
<math>\boxed{\textbf{(A)}}</math><br />
<br />
== Solution 3 ==<br />
<br />
If it's true for all values of <math>x</math> and <math>y</math>, then it's true for a specific value of <math>x</math> and <math>y</math>. Letting <math>x=5</math> and <math>y = 4</math> gives <math>P = 9</math> and <math>Q = 1</math>. The fraction becomes:<br />
<br />
<math>\frac{9 + 1}{9 - 1} - \frac{9 - 1}{9 + 1}</math><br />
<br />
<math>\frac{10}{8} - \frac{8}{10}</math><br />
<br />
<math>\frac{5}{4} - \frac{4}{5}</math><br />
<br />
<math>\frac{25}{20} - \frac{16}{20}</math><br />
<br />
<math>\frac{9}{20}</math><br />
<br />
Plugging in <math>(x, y) = (5, 4)</math> into the answer choices gives <math>\frac{9}{20}, \frac{9}{40}, 1, \frac{41}{20}, \frac{41}{40}</math>. Therefore, the answer is <math>\boxed{\textbf{(A)}}</math><br />
<br />
==See Also==<br />
{{AHSME 40p box|year=1964|num-b=3|num-a=5}}<br />
<br />
[[Category:Introductory Algebra Problems]]<br />
<br />
{{MAA Notice}}</div>
Talkinaway