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
$\endgroup$ 34 Answers
$\begingroup$Your solution is correct.
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: $|x-1|=x-1 \iff x \ge 1.$
$\endgroup$ $\begingroup$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
$\endgroup$ $\begingroup$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.
$\endgroup$ $\begingroup$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.
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