Question

The number of real solutions of $|x+3|-|4-x|=|8+x|$ is

Answer 1

There are $3$ turning points here: $-8,-3,4$
So we have $4$ cases:
$1)x<-8$
$\Rightarrow -(x+3)-(4-x)=-(8+x)\Rightarrow x=-1$
We reject the solution because our condition is not satisfied $(-1$ is not less than $-8$)
$2)-8\le x<-3$
$\Rightarrow -(x+3)-(4-x)=(8+x)\Rightarrow x=-15$
We reject the solution because our condition is not satisfied $(-15$ is not within $(-8,-3)$ interval.)
$3)-3\le x<4$
$\Rightarrow (x+3)-(4-x)=(8+x)\Rightarrow x=9$
We reject the solution because our condition is not satisfied ( $9$ is not within $(-3,4)$ interval.)
$4)x\ge 4$
$\Rightarrow (x+3)+(4-x)=(8+x)\Rightarrow x=-1$
We reject the solution because our condition is not satisfied $(-1$ is not more than $4$ )
Thus there is no solution.