# Thread: Equation involving modulus -- help..

1. ## Equation involving modulus -- help..

I need help with solving this equation:

|x - 1| = |x| + 1

I know that |x| = x when x >= 0 and |x| = -x when x < 0 but I am not sure how to apply it here.

When |x - 1| >= 0 , i.e. when x > 1, x - 1 = x + 1 => 0

Probably my reasoning is wrong thus far so I need help...

2. Hi

I can see 2 possibilities to answer
1) The equation is equivalent to |x-1| - |x| = 1
You can build a table like this
--------------------oo------------0-----------1------------+oo
-------|x-1|-------------1-x----------1-x----------x-1
--------|x|----------------x------------x-----------x
----|x-1|-|x|------------1------------1-2x----------1

Now you can easily conclude

2) An alternative way is to square the equation
|x-1| = |x| + 1
|x-1|² = (|x| + 1)²
x²-2x+1 = x²+2|x|+1 because |x|²=x²
-x = |x|

3. Hello, struck!

Solve: . $|x-1| \:=\:|x| + 1$
I'll do it the Long Way . . .

Since both $x-1$ and $x$ can be positive or negative,
. . there are four cases to consider:

[1] Both positive: . $x-1 > 0,\;\;x> 0 \quad\Rightarrow\quad x > 1$
. . .Then we have: . $x - 1 \:=\:x + 1 \quad\Rightarrow\quad 0 \:=\:2$ . . . impossible

[2] Positive-negative: . $x - 1 > 0,\;\;x < 0$
. . .But this means: . $(x > 1) \:\wedge\:(x < 0)$ . . . impossible

[3] Negative-positive: . $x-1 < 0,\;\;x > 0$
. . .Then we have: . $1 - x \:=\:x + 1 \quad\Rightarrow\quad x = 0$

[4] Negative-negative: . $x-1 < 0,\;\;x < 0 \quad\Rightarrow\quad x < 0$
. . .Then we have: . $1 - x \:=\:-x + 1 \quad\Rightarrow\quad 0 \:=\:0$ . . . always true.

The inequality is satisfied in cases [3] and [4]

The solution is: . $x \:\leq \:0$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

If we graph the two functions, we can "eyeball" the solution.

The graph of $y = |x|$ is a $\vee$, vertex at (0,0). .[1]

The graph of . $y \:=\:|x-1|$ . is [1], moved one unit to the right.
Code:
             \|
1*           /
|\         /
| \       /
|  \     /
|   \   /
|    \ /
- - - - + - - * - - - -
|     1

The graph of . $y \:=\:|x|+1$ .is [1], moved one unit upward.
Code:
              |
\     |     /
\    |    /
\   |   /
\  |  /
\ | /
\|/
1*
|
|
|
|
|
- - - - + - - - - - - -
|

Sketch them on the same coordinate system
. . and we can see the solotion: . $x \:\leq\:0$