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Q.
The number of integral roots of the equation $|x-1|+|x+2|-|x-3|=4$ is
Linear Inequalities
Solution:
$|x- 1|+ |x+ 2| -|x-3| =4$ , has three absolute value expressions, thus we divide the problem into four intervals :
(i) If $x< -2$ then
$-(x-1)- ( x+2)+ ( x-3)- =4 \Rightarrow x= -8$
(ii) If $-2 \le x < 1,$ then
-$(x-1)+1(x+2)+(x-3)=4$
$\Rightarrow x=4 \notin [-2, 1)$, hence rejected
(iii) If $1\le x < 3$, then
$(x-1)+(x+2)+(x-3)=4 \Rightarrow x=2$
(iv) If $x \ge 3$, then
$(x-1)+(x+2)-(x-3)=4 \Rightarrow x=0 \notin [3, \infty)$, hence rejected
$\therefore $ Solution set is $\left\{-8, 2\right\}$ and both are integers