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I am learning the topic of solving absolute value inequality question. I had mostly understood the steps in order to solve for an inequality. However, I'm still clueless of a step to solve the inequality below: Which is: Why does $ 3-x \ge 0$? I notice that 3-x is clearly not inside a radical, so it shouldn't have that requirement. Am I right?

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    $\begingroup$ Whoever wrote this "solution", their goal seems to be as confused as possible. The set of solutions is $x\geqslant2$, as seen by studying separately the cases $x\geqslant3$ and $x\lt3$. $\endgroup$ – Did Jul 22 '14 at 16:58
  • $\begingroup$ Just curious, what is the source for this "solution"? The more I look at it the more I dislike it. $\endgroup$ – mweiss Jul 22 '14 at 17:16
  • $\begingroup$ slideshare.net/Mrslily/… $\endgroup$ – Steven.Cooler Jul 23 '14 at 1:38
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Solve two separate cases.

Case 1: $x \ge 1$. Then solve $x-1 \ge 3-x$ to get $x \ge 2$.

Case 2: $x < 1$. Then solve $1-x \ge 3-x$ which is never true.

Hence the solution is $x \ge 2$.

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Squaring is not an equivalent transformation of an equation. I just would distinguish the two cases:

Case 1: $x\geq 1$

$x-1 \geq 3-x$

$2x \geq 4$

$x \geq 2$

Case 2: $x <1 $

$-x+1 \geq 3-x$

$1 \geq 3$

This is not true, no solution for case 2.

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  • $\begingroup$ You mean $x-1<0$ in case$\#2,$ right? $\endgroup$ – lab bhattacharjee Jul 22 '14 at 17:10
  • $\begingroup$ Yes, thanks for the hint. It was a typo. $\endgroup$ – callculus Jul 22 '14 at 17:15
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The requirement $3-x \geq 0$ is wrong. Whoever wrote this solution was probably thinking that $|x-1|$ is always non-negative, and $|x-1| \geq 3-x$, so $3-x$ is non-negative, but that reasoning is scrambled.

(If the inequality in the problem went in the other direction, i.e. if the original question were $|x-1| \leq 3-x$, then you could reason that $|x-1| \geq 0 \implies 3-x \geq 0$, but that would be a different problem.)

In fact you can just take any $x>3$ and see that indeed it is a solution to the inequality. For example, with $x=4$ we have $|4-1| \geq 3-4$, i.e. $3 \geq -1$, which is obviously true.


Edited to add: A better approach to solving the absolute value inequality is to break the problem into two cases.

  • Case 1: If $x-1$ is non-negative, then $|x-1| = x-1$, and so we have the compound inequality $x \geq 1$ and $x-1 \geq 3-x$. Solve and find $x \geq 2$.
  • Case 2: If $x-1$ is negative, then $|x-1| = 1-x$, and so we have the compound inequality $x < 1$ and $1-x \geq 3-x$. Solve and find this has no solutions.

So the solution is $x \geq 2$, full stop.

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There is no reason for $3-x \geq 0$. Unless it was part of assignment. See http://www.wolframalpha.com/input/?i=%7Cx-1%7C+%3E%3D+3-x.

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