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Difference between revisions of "1968 AHSME Problems/Problem 9"

(Created page with "== Problem == The sum of the real values of <math>x</math> satisfying the equality <math>|x+2|=2|x-2|</math> is: <math>\text{(A) } \frac{1}{3}\quad \text{(B) } \frac{2}{3}\quad...")
 
m (removed unnecessary case)
 
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== Solution ==
 
== Solution ==
−
<math>\fbox{}</math>
+
There are three cases: <math>\newline</math>
 +
(i) <math>(x+2), (x-2) \geq 0</math> <math>\newline</math>
 +
(ii) <math>x+2 \geq 0, x-2<0</math> <math>\newline</math>
 +
(iii) <math>(x+2), (x-2)<0</math> <math>\newline</math>
 +
In case (i), we solve the equation <math>x+2=2(x-2)</math> to get <math>x=6</math>. In case (ii), we solve the equation <math>x+2=2(2-x)</math> to get <math>x=\frac{2}{3}</math>. Case (iii) simply produces an equation which is that of case (i) multiplied on both sides by <math>-1</math>, so it yields the same solutions and thus may be discarded. Thus, if we add the real values of <math>x</math> which satisfy the original equation, we get <math>6+\frac{2}{3}=6\frac{2}{3}</math>, or answer choice <math>\fbox{E}</math>.
  
 
== See also ==
 
== See also ==
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{{AHSME box|year=1968|num-b=8|num-a=10}}   
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{{AHSME 35p box|year=1968|num-b=8|num-a=10}}   
  
 
[[Category: Introductory Algebra Problems]]
 
[[Category: Introductory Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 18:39, 17 July 2024

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 (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

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