## Thomas' Calculus 13th Edition

$x\in(-\infty,1]$
The absolute value is defined as $\quad |x|=\left\{\begin{array}{ll} x, & x\geq 0\\ -x, & x \lt 0 \end{array}\right.$ So, when $x-1\geq 0,\quad$ the LHS equals $x-1$, and when $x-1 \lt 0,\quad$ the LHS equals $-(x-1).$ We have two equations, one for each interval: $\left.\begin{array}{lclc|clc} \text{Int. } & x\geq 1 & & & x \lt 1 & \\ & & & & & \\ \text{Eq. } & x-1=1-x & /+x+1 & & -(x-1)=1-x & \\ & 2x=2 & /\div 2 & & -x+1=1-x & /+x-1\\ & x=1 & & & 0=0 & \end{array}\right.$ The solution for the interval $x\geq 1$ belongs to that interval, so it is a valid solution. The last equation for the interval $x \lt 1$ indicates that all numbers x from this interval satisfy the equation, so the whole interval is the solution set (for that interval). Combining solutions for both intervals, we have $x\in(-\infty,1]$