If $∣$ x - 1 $∣$ + $∣$ x + 1 $∣$ = 2, then find x.

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Solution

$| x - 1 | + | x + 1 | = 2$
Case 1 : $x \geq 1$
$| x - 1 | + | x + 1 | = 2$
$\Rightarrow x - 1 + x + 1 = 2$
$\Rightarrow 2 x = 2$
$x = 1$
Case 2 : $- 1 \leq x < 1$
$| x - 1 | + | x + 1 | = 2$
$\Rightarrow 1 - x + x + 1 = 2$
$\Rightarrow 2 = 2$
$∵$ LHS=RHS
$∴$ $| x - 1 | + | x + 1 | = 2$ is true for all $x ϵ [- 1, 1]$
Case 3 : $x < - 1$
$1 - x + - 1 - x = 2$
$\Rightarrow - 2 x = 2$
$x = - 1$
But the image is $x < - 1$
$∴$ the values of $x$ satisfying
$| x - 1 | + | x + 1 | = 2$
lies in the range $[- 1, 1]$

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