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 $a=-b$)
$x+3=2x+1 \Rightarrow x=2$
$-(x+3)=2x+1 \Rightarrow x=-\dfrac 4 3$
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
Case 2: If x < 0. then
So we can conclude that the solution to the system is at x=2?
I checked Wolfram but they have a solution of http://www.wolframalpha.com/input/?i...3D%7C2x%2B1%7C
Any Help is appreciated.
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 $a=-b$)
$x+3=2x+1 \Rightarrow x=2$
$-(x+3)=2x+1 \Rightarrow x=-\dfrac 4 3$
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.
Personally, I would not look a those cases. |x+ 3|= |2x+ 1| immediately gives x+ 3= 2x+ 1 and x+ 3= -(2x+ 1).
Case 2: If x < 0. then
So we can conclude that the solution to the system is at x=2?
I checked Wolfram but they have a solution of |x+3|=|2x+1| - Wolfram|Alpha
Any Help is appreciated.