How to solve Absolute Value Inequality: |x-1| ≥ 3-x 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?

 A: 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 $x \ge 2$.
A: 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.
A: 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.
A: 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.
