solve
|x-1|=1-x.
i got it down to x=1 but it the solution can also be any number between 0 and 1. help???
First of all we have the condition $\displaystyle 1-x\geq 0$ (because the left side is $\displaystyle \geq 0$). So, $\displaystyle x\leq 1$.
But, for $\displaystyle x\leq 1$ we have $\displaystyle |x-1|=1-x$.
So, the inequality stands for all $\displaystyle x\in(-\infty,1]$.
You have only to understand this concept:
If x >= 0, then |x| = x -- The absolute value does nothing to a nonnegative value.
If x < 0, then |x| = -x -- Example, |-3| = -(-3) = 3
So, you have
|x-1|
For x-1 >= 0, or x >= 1, |x-1| = x-1. Solve x-1 = 1-x, but ONLY for x >= 1
For x-1 < 0, or x < 1, |x-1| = -(x-1) = 1-x. Solve 1-x = 1-x, but ONLY for x < 1
Let's see what you get.
$\displaystyle |x|=x$ if $\displaystyle x\geq 0$ and $\displaystyle |x|=-x$ is $\displaystyle x<0$ by definition.
Given $\displaystyle |x-1| = |y|$ where $\displaystyle y=x-1$. We have $\displaystyle |y| = y$ if $\displaystyle y\geq 0$ meaning $\displaystyle |x-1| = x-1$ if $\displaystyle x-1\geq 0$ so this means $\displaystyle |x-1|=x-1$ if $\displaystyle x\geq 1$.
Similaryl $\displaystyle |x-1|=1-x$ if $\displaystyle x<1$.