# 1954 AHSME Problems/Problem 22

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

## Problem 22

The expression $\frac{2x^2-x}{(x+1)(x-2)}-\frac{4+x}{(x+1)(x-2)}$ cannot be evaluated for $x=-1$ or $x=2$, since division by zero is not allowed. For other values of $x$:

$\textbf{(A)}\ \text{The expression takes on many different values.}\\ \textbf{(B)}\ \text{The expression has only the value 2.}\\ \textbf{(C)}\ \text{The expression has only the value 1.}\\ \textbf{(D)}\ \text{The expression always has a value between }-1\text{ and }+2.\\ \textbf{(E)}\ \text{The expression has a value greater than 2 or less than }-1.$

## Solution

$\frac{2x^2-x}{(x+1)(x-2)}-\frac{4+x}{(x+1)(x-2)} = \frac{2x^2-2x-4}{(x+1)(x-2)}$

This can be factored as $\frac{(2)(x^2-x-2)}{(x+1)(x-2)} \implies \frac{(2)(x+1)(x-2)}{(x+1)(x-2)}$, which cancels out to $2 \implies \boxed{\textbf{(B)} \text{The expression has only the value 2}}$.

 1954 AHSC (Problems • Answer Key • Resources) Preceded byProblem 21 Followed byProblem 23 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