Answer
$x\in(-\infty,1]$
Work Step by Step
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
$x\in(-\infty,1]$
