https://artofproblemsolving.com/wiki/index.php?title=1954_AHSME_Problems/Problem_22&feed=atom&action=history1954 AHSME Problems/Problem 22 - Revision history2024-03-28T17:38:36ZRevision history for this page on the wikiMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=1954_AHSME_Problems/Problem_22&diff=85233&oldid=prevXiej: Created page with "== Problem 22== The expression <math>\frac{2x^2-x}{(x+1)(x-2)}-\frac{4+x}{(x+1)(x-2)}</math> cannot be evaluated for <math>x=-1</math> or <math>x=2</math>, since division by..."2017-04-15T22:12:59Z<p>Created page with "== Problem 22== The expression <math>\frac{2x^2-x}{(x+1)(x-2)}-\frac{4+x}{(x+1)(x-2)}</math> cannot be evaluated for <math>x=-1</math> or <math>x=2</math>, since division by..."</p>
<p><b>New page</b></p><div>== Problem 22==<br />
<br />
The expression <math>\frac{2x^2-x}{(x+1)(x-2)}-\frac{4+x}{(x+1)(x-2)}</math> cannot be evaluated for <math>x=-1</math> or <math>x=2</math>, <br />
since division by zero is not allowed. For other values of <math>x</math>: <br />
<br />
<math> \textbf{(A)}\ \text{The expression takes on many different values.}\\ \textbf{(B)}\ \text{The expression has only the value 2.}\\ \textbf{(C)}\ \text{The expression has only the value 1.}\\ \textbf{(D)}\ \text{The expression always has a value between }-1\text{ and }+2.\\ \textbf{(E)}\ \text{The expression has a value greater than 2 or less than }-1. </math><br />
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== Solution ==<br />
<math>\frac{2x^2-x}{(x+1)(x-2)}-\frac{4+x}{(x+1)(x-2)} = \frac{2x^2-2x-4}{(x+1)(x-2)}</math><br />
<br />
This can be factored as <math>\frac{(2)(x^2-x-2)}{(x+1)(x-2)} \implies \frac{(2)(x+1)(x-2)}{(x+1)(x-2)}</math>, which cancels out to <math>2 \implies \boxed{\textbf{(B)} \text{The expression has only the value 2}}</math>.<br />
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{{AHSME 50p box|year=1954|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Xiej