Question 1184798
<br>
For x<-1, {{{abs(x+1)=-x-1}}}; for x>-1, {{{abs(x+1)=x+1}}}.<br>
For x<3/2, {{{abs(2x-3)=-2x+3}}}; for x>3/2< {{{abs(2x-3)=2x-3}}}.<br>
To solve the equation then, we need to consider three intervals of possible values for x: (1) x less than -1; (2) x between -1 and 3/2; and (3) x greater than 3/2.<br>
Solve the equation on each of those intervals, remembering that any "solution" we get has to be in the interval we are working in.<br>
(1) x less than -1<br>
{{{abs(x+1)+abs(2x-3)=-x-1-2x+3}}}<br>
{{{-x-1-2x+3=8}}}
{{{-3x+2=8}}}
{{{-3x=6}}}
{{{x=-2}}}<br>
Since our solution x=-2 is in the interval x<-1, it is a solution; we can easily check that:<br>
{{{abs(x+1)+abs(2x-3)=abs(-1)+abs(-7)=1+7=8}}}<br>
(2) x between -1 and 3/2<br>
{{{abs(x+1)+abs(2x-3)=x+1-2x+3}}}<br>
{{{x+1-2x+3=8}}}
{{{-x=4}}}
{{{x=-4}}}<br>
That solution is NOT in the interval we are working in; it is not a solution.<br>
(3) x greater than 3/2<br>
{{{abs(x+1)+abs(2x-3)=x+1+2x-3}}}<br>
{{{x+1+2x-3=8}}}
{{{3x=10}}}
{{{x=10/3}}}<br>
That solution is in the interval we are working in, so it is a solution.<br>
{{{abs(x+1)+abs(2x-3)=abs(13/3)+abs(11/3)=13/3+11/3=24/3=8}}}<br>
ANSWERS: x=-2; x=10/3<br>
A graph, showing abs(x+1)+abs(2x-3) (red) = 8 (green) at x=-2 and at x=10/3....<br>
{{{graph(400,400,-5,5,-2,10,abs(x+1)+abs(2x-3),8)}}}<br>