https://artofproblemsolving.com/wiki/index.php?title=1977_AHSME_Problems/Problem_6&feed=atom&action=history
1977 AHSME Problems/Problem 6 - Revision history
2024-03-19T03:37:05Z
Revision history for this page on the wiki
MediaWiki 1.31.1
https://artofproblemsolving.com/wiki/index.php?title=1977_AHSME_Problems/Problem_6&diff=160475&oldid=prev
Jiang147369: Created page with "== Problem 6 == If <math>x, y</math> and <math>2x + \frac{y}{2}</math> are not zero, then <math>\left( 2x + \frac{y}{2} \right)^{-1} \left[(2x)^{-1} + \left( \frac{y}{2} \rig..."
2021-08-17T18:57:23Z
<p>Created page with "== Problem 6 == If <math>x, y</math> and <math>2x + \frac{y}{2}</math> are not zero, then <math>\left( 2x + \frac{y}{2} \right)^{-1} \left[(2x)^{-1} + \left( \frac{y}{2} \rig..."</p>
<p><b>New page</b></p><div>== Problem 6 ==<br />
<br />
If <math>x, y</math> and <math>2x + \frac{y}{2}</math> are not zero, then<br />
<math>\left( 2x + \frac{y}{2} \right)^{-1} \left[(2x)^{-1} + \left( \frac{y}{2} \right)^{-1} \right]</math><br />
equals<br />
<br />
<math>\textbf{(A) }1\qquad<br />
\textbf{(B) }xy^{-1}\qquad<br />
\textbf{(C) }x^{-1}y\qquad<br />
\textbf{(D) }(xy)^{-1}\qquad <br />
\textbf{(E) }\text{none of these} </math> <br />
<br />
<br />
== Solution ==<br />
We can write <math>\left( 2x+ \frac{y}{2} \right)^{-1} = \left( \frac{4x+y}{2} \right)^{-1} = \frac{2}{4x+y}</math>.<br />
<br />
Then, the expression simplifies to <cmath>\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}.</cmath><br />
<br />
<br />
Thus, our answer is <math>\textbf{(D) }(xy)^{-1}</math>. ~[[User:Jiang147369|jiang147369]]<br />
<br />
<br />
==See Also==<br />
{{AHSME box|year=1977|num-b=5|num-a=7}}<br />
{{MAA Notice}}</div>
Jiang147369