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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

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

4 Answers 4

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

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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

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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.

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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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  • $\begingroup$ Please don't add "thank you" as an answer. Instead, vote up the answers that you find helpful. - From Review $\endgroup$
    – MathAdam
    Jan 25, 2019 at 19:53
  • $\begingroup$ @Adam: I am not the OP. I answered the OP's question, and edited the question's tags. If you are going to review things, please be more careful in the future. $\endgroup$ Jan 26, 2019 at 14:19
  • $\begingroup$ When I reviewed the answer, it said exactly this: "Nicely done! That's about the most straightforward approach you could take." It's been edited since. $\endgroup$
    – MathAdam
    Jan 26, 2019 at 16:06
  • $\begingroup$ True. At no point did it say "Thank you!" $\endgroup$ Jan 26, 2019 at 16:07
  • $\begingroup$ Indeed. Canned response. You said "Nicely done." Instead of undeleting your non-answer and then adding an answer, why not just add a new answer? $\endgroup$
    – MathAdam
    Jan 26, 2019 at 16:12

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