$\left|x-1\right|=1-x$ solve the equation.

Question

$\left|x-1\right|=1-x$

$\left|x-1\right| \Rightarrow \{ x-1, x\geq1; -x+1, x\lt1 \}$

When $x\geq1, x-1=1-x \Rightarrow 2x=2 \Rightarrow x=1$

When $x\lt1, -(x-1)=1-x \Rightarrow -x+1=1-x \Rightarrow 0=0$ so, true for all x

All I can do is just this.

I can't go any further.

Answer 1

For real $a$ we have

$$a=|a| \iff a \ge 0.$$

Since $|x-1|=|1-x|$, we get

$$|x-1|=1-x \iff |1-x|=1-x \iff 1-x \ge 0 \iff x \le 1.$$

Answer 2

$|1-x|=|x-1|=1-x$;

$y:=1-x$; Then

$|y|=y$, or $y \ge 0$;

(Recall the definition of the abs. value)

$y=1-x \ge 0$;

$x \le 1$.

Answer 3

$$ case1: x>1 \implies|x-1| = x-1=1-x, so x =1$$ $$ case 2: x<1 \implies|x-1|=1-x = RHS$$ $$\therefore$x= 1$\cup all x <1 $$

PS: Draw the graph of LHS and RHS to see it more clearly.