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 $\fbox{E}$ .

1968 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

1968 AHSME Problems/Problem 9

Problem

Solution

See also