1968 AHSME Problems/Problem 9

Problem

The sum of the real values of $x$ satisfying the equality $|x+2|=2|x-2|$ is:

$\text{(A) } \frac{1}{3}\quad \text{(B) } \frac{2}{3}\quad \text{(C) } 6\quad \text{(D) } 6\tfrac{1}{3}\quad \text{(E) } 6\tfrac{2}{3}$

Solution

There are four cases: $\newline$ (i) $(x+2), (x-2) \geq 0$ $\newline$ (ii) $x+2 \geq 0, x-2<0$ $\newline$ (iii) $(x+2), (x-2)<0$ $\newline$ (iv) $x+2<0, x-2 \geq 0$ . $\newline$ In case (i), we solve the equation $x+2=2(x-2)$ to get $x=6$ . In case (ii), we solve the equation $x+2=2(2-x)$ to get $x=\frac{2}{3}$ . Cases (iii) and (iv) simply produce equations which are those of cases (i) and (ii), respectively, multiplied on both sides by $-1$ , so they yield the same solutions and thus may be discarded. Thus, if we add the real values of $x$ which satisfy the original equation, we get $6+\frac{2}{3}=6\frac{2}{3}$ , or answer choice $\fbox{E}$ .

1968 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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1968 AHSME Problems/Problem 9

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