Solve the absolute value inequality |3x - 1| - |x + 1| > 0 Below I will post my attempt.
Breaking up both arguments into cases: Case 1: (3x - 1) ≥ 0 . This is when x ≥ 1/3 (3x - 1) - (x + 1) > 0 3x - 1 - x - 1 > 0 2x - 2 > 0 2x > 2 x > 1 Case 2: (3x - 1) < 0. This is when x < 1/3 -(3x - 1) - -(x + 1) > 0 -3x + 1 -(-x - 1) > 0 -3x + 1 + x + 1 > 0 -2x + 2 > 0 -2x > -2 x < 1 Case 3: (x + 1) ≥ 0. This is when x ≥ -1 (3x - 1) - (x + 1) > 0 3x - 1 -x -1 > 0 2x - 2 > 0 2x > 2 x > 1 [i]But, what if for the sake of argument, you pick x = 0 within that interval [-1,∞)[/i] -(3x - 1) - (x + 1) > 0 -3x + 1 -x - 1 > 0 -4x + 0 > 0 x > 0 Case 4: (x + 1) < 0. This is when x < -1 -(3x - 1) - -(x + 1) > 0 -3x + 1 - (-x -1) > 0 -3x + 1 + x + 1 > 0 -2x + 2 > 0 -2x > -2 x < 1 I am not sure as to whether or not I did this properly at all.
I get x<0 , x>1
Well, at least I was right about it being only two possibilities
Ok how did you get only those two solutions?
Squaring both sides. ; )
Maybe I mean that is How I got but Zarkon may have some different method.
The problem with squaring is that you sometimes have wrong solutions in your answer so every time you square you must check your main function with the solutions you get.
@ hero put 1/2 into the original equation
How are you able to square both sides?
What I mean is I added | x + 1| to both sides first
\[|3x - 1| - |x + 1| > 0\] \[|3/2 - 1| - |1/2 + 1| > 0\] \[1/2-3/2>0\] \[-1>0\]
Nope.... what I wrote is still correct
Oh, I made a mistake
the answer is x<0, x>1
I thought it was |3x+1|
You don't have to square both sides to solve these though
I didn't do it that way
ah yeah you are correct but I have to get I mean right now I am not perfect through the other way so I square as I am more confident in the squaring method.
But I will leave squaring as soon as I get good in the other method : )
I don't really trust squaring
You get extraneous roots.
The other way splits it into four inequalities with potentially four possible solutions
Sometimes you only get two solutions, like in this case
I know but I need more practice at that.
squaring in this case is fine if you use equality then pick numbers in your intervals to see if they satisfy the original equation
I need more practice at just inputting the correct equation. I used |3x+1| instead of |3x-1|
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