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 three cases: $\newline$ (i) $(x+2), (x-2) \geq 0$ $\newline$ (ii) $x+2 \geq 0, x-2<0$ $\newline$ (iii) $(x+2), (x-2)<0$ $\newline$ 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 $x=\frac{2}{3}$. Case (iii) simply produces an equation which is that of case (i) multiplied on both sides by $-1$, so it yields 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 $\fbox{E}$.

See also

1968 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
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