1. ## Use of modulus..

I have tried a lot of questions on Modulus...yet it creates trouble for me when it comes to what would be the answer to a questions like these:
$||x-2|-2|=x$ find value of x
and one more, $|3-|2x+5||=x+2$ find $x$..

answer to the first one is $0<=x<=2$ and of second one is 0, -4/3

I will be thankful to any sort of help.

2. Originally Posted by findmehere.genius
I have tried a lot of questions on Modulus...yet it creates trouble for me when it comes to what would be the answer to a questions like these:
$||x-2|-2|=x$ find value of x
and one more, $|3-|2x+5||=x+2$ find $x$..

answer to the first one is $0<=x<=2$ and of second one is 0, -4/3

I will be thankful to any sort of help.
Hi

$||x-2|-2|=x$ find value of x

Since there is |x-2| you need to consider 2 cases :
1) $x\leq 2$
In this case |x-2| = 2-x (because $x-2 \leq 0$)
The equation becomes |-x|=x or |x|=x (because |-x|=|x|)
which means $x\geq 0$
The solutions in this case are therefore $0 \leq x \leq 2$

2) $x\geq 2$
In this case |x-2| = x-2 (because $x-2 \geq 0$)
The equation becomes |x-4|=x

2 sub-cases :
2a) $x\leq 4$
The equation becomes 4-x=x which means x=2
2b) $x\geq 4$
The equation becomes x-4=x which has no solution
The solutions in this case are therefore $x = 2$

The overall solutions are therefore $0 \leq x \leq 2$

3. The method is the same for the second one

$|3-|2x+5||=x+2$ find $x$

Case 1 : $x \leq -\frac52$
The equation becomes $|3-(-2x-5)|=x+2$ or $|2x+8|=x+2$

Case 1a : $x \leq -4$
The equation becomes $-(2x+8)=x+2$ or $x = -\frac{10}{3}$
But $-\frac{10}{3} > -4$ therefore there is no solution for this case

Case 1b : $x \geq -4$
The equation becomes $2x+8=x+2$ or $x = -6$
But $-6 < -4$ therefore there is no solution for this case

Case 2 : $x \geq -\frac52$
The equation becomes $|3-(2x+5)|=x+2$ or $|-2x-2|=x+2$

Case 2a : $x \leq -1$
The equation becomes $-2x-2=x+2$ or $x = -\frac43$
And $-\frac43 \leq -1$ therefore $-\frac43$ is one solution

Case 2b : $x \geq -1$
The equation becomes $2x+2=x+2$ or $x = 0$
And $0 \geq -1$ therefore $0$ is one solution

Finally the set of solutions is -4/3 and 0

4. Hello, findmehere.genius!

I graphed the first one . . . Got the same answer as running-gag.

$\bigg||x-2|-2\bigg|\:=\:x$ . Find value of $x.$
I'll graph it in baby steps . . .

We know what $y \:=\:x$ looks like:
Code:
                |
|     *
|   *
| *
- - - - - * - - - - -
* |
*   |
*     |
|

Then we have: . $y \:=\:|x|$
The absolute value says: anything below the x-axis is reflected upward.
Code:
                |
*       |       *
*     |     *
*   |   *
* | *
- - - - - * - - - - -
|
|

Then $y \:=\:|x-2|$ moves the graph 2 units to the right.
Code:
        *   |           *
* |         *
*       *
| *   *
- - - + - * - - - - -
|   2
|

Then $y \:=\:|x-2| -2$ moves the graph down 2 units.
Code:
            |
*     |             *
*   |           *
* |         *
- - - * - - - * - - -
| *   *
|   *
|

Finally, $y \:=\:||x-2| - 2|$
. . Anything below the $x$-axis is reflected upward.
Code:
            |
*     |             *
*   |   *       *
* | *   *   *
- - - * - + - * - - - -
|   2
|
There!

What does this graph have in common with $y = x$ ?
. . (See the first graph.)

We find that the graphs coincide for: . $0 \:\leq\:x\:\leq\:2$