# Solve the equation |x-1|=x-1

Solve the equation:$$|x-1|=x-1$$

My solution:

Case 1 :$$x\ge1$$, Hence $$x-1=x-1$$, therefore infinite solution

Case 2 :$$x<1$$, Hence $$1-x=x-1$$,$$x=1$$, hence no solution

But the solution i saw concept used is $$x\le1$$ in lieu of $$x<1$$

Hence final answer is $$[1,\infty]$$, is this concept correct

• Seems correct to me. Jan 25, 2019 at 11:57
• Take a case |x-a|=1 Jan 25, 2019 at 12:02
• Slight error... make the right bracket (right of the $\infty$) a right parenthesis. In other words, the solution set should be $$[1,\infty)$$ Jan 25, 2019 at 12:13

My solution: if $$|x-1|=x-1$$, then $$x-1 \ge 0$$, hence $$x \ge 1$$. For $$x \ge 1$$ your equations reads as follows: $$x-1=x-1$$.

Hence: $$|x-1|=x-1 \iff x \ge 1.$$

If you were to make case 1 cover $$x \gt 1$$ and case 2 cover $$x \le 1$$ then you would get a similar answer for case 1, i.e. all such $$x \in (1,\infty)$$ satisfy the equation

while for case 2: $$\qquad1-x=x-1 \implies x=1$$

and your combined solution would then be $$x \in (1,\infty) \cup \{1\} = [1,\infty)$$, the same solution as your original method

Your answer is right, apart from the square bracket pointed out in @ElevenEleven's comment ($$\infty$$ can't be the upper end of a closed interval).

Another way to get it is to note that $$|a|=a$$ only when $$a\geq 0$$, so for $$a=x-1$$,

$$|x-1|=x-1$$ implies $$x-1\geq 0$$

which gives you the answer without needing to consider different cases.

Nicely done! Ultimately, you don't need to consider the second case, since $$\lvert t\rvert\neq t$$ for $$t<0,$$ but it doesn't hurt to be thorough. The only thing to change about your conclusion is from $$[1,\infty]$$ to $$[1,\infty),$$ as "$$\infty$$" is a notational convention and not a real number.