1. ## Absolute value equation

$|x+3| = |2x+1|$

How do we go about solving this? I know there are two cases when x>0 and when x<0, but how do we set it up afterwards?

Edit: I may have solved this:

Case 1: If x > 0, then

$x+3 = 2x +1$
$x=2$

Case 2: If x < 0. then
$-(x+3) = - (2x+1)$
$x=2$

So we can conclude that the solution to the system is at x=2?

I checked Wolfram but they have a solution of $\frac{-4}{3}$ http://www.wolframalpha.com/input/?i...3D%7C2x%2B1%7C

Any Help is appreciated.

2. ## Re: Absolute value equation

x being greater or less than zero isn't of issue here.

there are 2 ways that $|a|=|b|$

$a=b$ (this is the same as $-a=-b$)

$-a=b$ (this is the same as $a=-b$)

$x+3=2x+1 \Rightarrow x=2$

$-(x+3)=2x+1 \Rightarrow x=-\dfrac 4 3$

3. ## Re: Absolute value equation

So we have two solutions at x= 2 and -4/3 ?

4. ## Re: Absolute value equation

Originally Posted by sakonpure6
$|x+3| = |2x+1|$

How do we go about solving this? I know there are two cases when x>0 and when x<0, but how do we set it up afterwards?

Edit: I may have solved this:

Case 1: If x > 0, then
No, that's not true. the cases for |a| are a> 0 and a< 0. Here, you have |x+ 3| and |2x+ 1| so the cases, for |x+ 3| are x+ 3> 0 and x+ 3< 0 which give x>-3 and x< -3. The "cut point" for x is -3, not 0. For |2x+ 1|, the cases are 2x+1> 0 and 2x+ 1< 0 which give x> -1/2 and x< -1/2. The "cut point" for x is -1/2.

So there are really three cases: x< -3, -3< x< -1/2, and x> -1/2.

$x+3 = 2x +1$
$x=2$

Case 2: If x < 0. then
$-(x+3) = - (2x+1)$
$x=2$

So we can conclude that the solution to the system is at x=2?

I checked Wolfram but they have a solution of $\frac{-4}{3}$ |x+3|=|2x+1| - Wolfram|Alpha

Any Help is appreciated.
Personally, I would not look a those cases. |x+ 3|= |2x+ 1| immediately gives x+ 3= 2x+ 1 and x+ 3= -(2x+ 1).

5. ## Re: Absolute value equation

Originally Posted by sakonpure6
So we have two solutions at x= 2 and -4/3 ?
yes

6. ## Re: Absolute value equation

Originally Posted by sakonpure6
$|x+3| = |2x+1|$

How do we go about solving this?
Here is a different way to do these.
$a^2\le\ b^2~\iff~|a|\le|b|~.$ So solve $(x+3)^2=(2x+1)^2$