What is the Solution for |x+1| = 2x-1?

[Rusho]

Is this right:

x+1 = -2x+1
2x+x = 1-1
3x = 0

No solution?

 

[VietDao29]

Is this right:

x+1 = -2x+1
2x+x = 1-1
3x = 0

No solution?

No, that's not correct.
I'll give you some hints to start of the problems.
First, you should notice that, can |A| ever be negative?
So if you have an equation: |A| = B, can B be negative?
And you will have:
|A| = \left\{ \begin{array}{l} A, \quad \mbox{if } A \geq 0 \\ -A , \quad \mbox{if } A < 0 \end{array} \right.
So to solve something like |A| = B, you will have to solve the system of equations:
\left\{ \begin{array}{l} B \geq 0 \\ \left[ \begin{array}{l} A = B \\ A = -B \end{array} \right. \end{array} \right.
Can you get it? Can you go from here? :)

 

[Rusho]

ok so
-x+1 = 2x-1
-3x = -2
x=-2/3

?

 

[VietDao29]

ok so
-x+1 = 2x-1
-3x = -2
x=-2/3

?

Again, it's wrong...
First -(x + 1) is -x - 1, not -x + 1.
Now, just do it step by step, and see if you can get the answer.
First, in the equation you have:
|x + 1| = 2x - 1.
Just compare it with |A| = B, so what's A, and what's B, you think?
And can you solve this system
\left\{ \begin{array}{l} B \geq 0 \\ \left[ \begin{array}{l} A = B \\ A = -B \end{array} \right. \end{array} \right.?
But anyway, you should read my previous post again to understand the concept more deeply.
Can you go from here? :)

 

[Rusho]

so |A| is |x+1| and B is 2x-1
|A| can be negative
B cannot
-x-1 = 2x-1
-2x-x = 1-1
-3x = 0 ?

I'm sorry if I'm just not seeing this

 

[VietDao29]

so |A| is |x+1| and B is 2x-1
|A| can be negative
B cannot
-x-1 = 2x-1
-2x-x = 1-1
-3x = 0 ?

I'm sorry if I'm just not seeing this

The absolute value of A is never negative.
Hence |A| >= 0
If |A| = B, and A >= 0, it's true that B >= 0, right? Now, just look at my first post in this thread to see if you can get it.
Now can you solve for the value of x that makes B non-negative?
By the way, there should be some examples in your books, let's give it a glance to see if you find any.
I'll give you an example, though:
------------------
Example:
Solve the equation:
|2x + 1| = x - 1
\Leftrightarrow \left\{ \begin{array}{l} x - 1 \geq 0 \\ \left[ \begin{array}{l} 2x + 1 = x - 1 \\ 2x + 1 = -(x - 1) \end{array} \right. \end{array} \right.
\Leftrightarrow \left\{ \begin{array}{l} x \geq 1 \\ \left[ \begin{array}{l} x = - 2 \\ 2x + 1 = -x + 1 \end{array} \right. \end{array} \right.
\Leftrightarrow \left\{ \begin{array}{l} x \geq 1 \\ \left[ \begin{array}{l} x = - 2 \\ 3x = 0 \end{array} \right. \end{array} \right.
\Leftrightarrow \left\{ \begin{array}{l} x \geq 1 \\ \left[ \begin{array}{l} x = - 2 \\ x = 0 \end{array} \right. \end{array} \right.
x = -2 is not a valid solution, right? Because in order for x - 1 >= 0, x must be greater than or equal to 1, and -2 is obviously less than 1.
And neither is x = 0, hence there's no x such that:
|2x + 1| = x - 1
------------------
Can you go from here? :)

 

[Rusho]

alright, I'll work on it, thanks for your help!

 

[krab]

|x+1| is always >0. Now look at the right side. It means that x MUST be larger than 1/2. So x+1 is >0 and we drop the absolute value sign: x+1=2x-1. It's pretty easy to solve.