Question 1207733
<br>
{{{abs(x+abs(3x-2))=2}}}<br>
If we think of this as a function, we know the behavior of the graph changes when (3x-2) is 0 -- i.e., at x=2/3.  So let's solve two separate absolute value equations, considering only values of x greater than or equal to 2/3 in one equation and considering only values of x less than 2/3 in the other.<br>
(1) If x is greater than or equal to 2/3, then |3x-2| is equal to 3x-2:<br>
{{{abs(x+(3x-2))=2}}}
{{{abs(4x-2)=2}}}<br>
(1a) {{{4x-2=2}}}
{{{4x=4}}}
{{{x=1}}}<br>
x=1 satisfies our condition that x is greater than or equal to 2/3, so x=1 is a solution.<br>
OR<br>
(1b) {{{4x-2=-2}}}
{{{4x=0}}}
{{{x=0}}}<br>
x=0 does NOT satisfy our condition that x is greater than or equal to 2/3, so x=0 is not a solution.<br>
(2) If x is less than 2/3, then |3x-2| is equal to -(3x-2) = -3x+2:<br>
{{{abs(x+(-3x+2))=2}}}
{{{abs(-2x+2)=2}}}<br>
(2a) {{{-2x+2=2}}}
{{{-2x=0}}}
{{{x=0}}}<br>
x=0 satisfies our condition that x is less than 2/3, so x=0 is a solution.<br>
OR<br>
(2b) {{{-2x+2=-2}}}
{{{-2x=-4}}}
{{{x=2}}}<br>
x=2 does NOT satisfy our condition that x is less than 2/3, so x=2 is not a solution.<br>
We found two solutions above, in cases (1a) and (2a).<br>
ANSWER: x=1 or x=0<br>