Question 1148537
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{{{2x+1=0}}} when x = -1/2.  So consider two cases, one with x < -1/2 and the other with x > -1/2.<br>
Case 1: x < -1/2; 2x+1 < 0<br>
{{{abs(2x+1) = -2x-1}}}
{{{abs(x-abs(2x+1)) = abs(x-(-2x-1)) = abs(3x+1) = -3x-1}}}<br>
The solution on the interval x < -1/2 is then<br>
{{{-3x-1 = 3}}}
{{{-3x = 4}}}
{{{x = -4/3}}}<br>
Case 2: x > -1/2; 2x+1 > 0<br>
{{{abs(2x+1) = 2x+1}}}
{{{abs(x-abs(2x+1)) = abs(x-(2x+1)) = abs(-x-1) = x+1}}}<br>
The solution on the interval x > -1/2 is then<br>
{{{x+1 = 3}}}
{{{x = 2}}}<br>
ANSWER: There are two solutions to the equation: -4/3 and 2.<br>
A graph, showing the intersection of the two graphs at x=-4/3 and x=2....<br>
{{{graph(400,200,-3,3,-1,5,abs(x-abs(2x+1)),3)}}}<br>