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

score 3 · Accepted Answer · answered Sep 06 '19 at 07:09

3

For real $a$ we have

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

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

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

answered Sep 06 '19 at 07:09

Fred

Oh, I can't understand it. Would not the equation $\left|x-1\right|=\left|1-x\right|$ is true when the expression $x-1\geq0$? – M. Çağlar TUFAN Sep 06 '19 at 07:18
1

@ Muhammed Ç. TUFAN, its always the same and $\left|a-b\right|=\left|b-a\right|$, for example take $\left|2-3\right|=\left|3-2\right|$. – Sep 06 '19 at 07:26
@Grothendieck Is this $\left|a-b\right|=\left|b-a\right|$ expression true for all conditions? I tried to use some different conditions but I they out the same result. Note: After I think about it a little more, I figured out that distance between two things never changes. Thanks! – M. Çağlar TUFAN Sep 06 '19 at 07:35

score 1 · Answer 2 · answered Sep 06 '19 at 07:42

1

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

$y:=1-x$; Then

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

(Recall the definition of the abs. value)

$y=1-x \ge 0$;

$x \le 1$.

answered Sep 06 '19 at 07:42

Peter Szilas

Thank you for the explanation @PeterSzilas! I got the idea why we are able to $1-x\geq0$. – M. Çağlar TUFAN Sep 06 '19 at 07:46
1

Muhammed.Welcome. – Peter Szilas Sep 06 '19 at 07:49

George carlin · Answer 3 · 2019-09-06T09:02:40.617

1

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

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

edited Sep 06 '19 at 09:02

answered Sep 06 '19 at 07:59

George carlin

3 Answers3