1. Absolute Value

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???

2. First of all we have the condition $1-x\geq 0$ (because the left side is $\geq 0$). So, $x\leq 1$.
But, for $x\leq 1$ we have $|x-1|=1-x$.
So, the inequality stands for all $x\in(-\infty,1]$.

3. Originally Posted by scheuerman
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???
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.

4. Thank you guys soo much... but i still really dont get it...

5. Originally Posted by scheuerman
Thank you guys soo much... but i still really dont get it...
$|x|=x$ if $x\geq 0$ and $|x|=-x$ is $x<0$ by definition.

Given $|x-1| = |y|$ where $y=x-1$. We have $|y| = y$ if $y\geq 0$ meaning $|x-1| = x-1$ if $x-1\geq 0$ so this means $|x-1|=x-1$ if $x\geq 1$.

Similaryl $|x-1|=1-x$ if $x<1$.