Question

Solve the absolute value inequality
|3x - 1| - |x + 1| > 0

Below I will post my attempt.

Answer 1

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.

Answer 2

I get x<0 , x>1

Answer 3

Well, at least I was right about it being only two possibilities

Answer 4

Ok how did you get only those two solutions?

Answer 5

Squaring both sides. ; )

Answer 6

Maybe I mean that is How I got but Zarkon may have some different method.

Answer 7

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.

Answer 8

@ hero put 1/2 into the original equation

Answer 9

How are you able to square both sides?

Answer 10

What I mean is I added | x + 1| to both sides first

Answer 11

\[(3x-1)^2=(x+1)^2\Rightarrow x=0,1\]

Answer 12

\[|3x - 1| - |x + 1| > 0\]

\[|3/2 - 1| - |1/2 + 1| > 0\]
\[1/2-3/2>0\]

\[-1>0\]

Answer 13

Nope.... what I wrote is still correct

Answer 14

really ?

Answer 15

Oh, I made a mistake

Answer 16

the answer is x<0, x>1

Answer 17

I thought it was |3x+1|

Answer 18

craps

Answer 19

You don't have to square both sides to solve these though

Answer 20

I didn't do it that way

Answer 21

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.

Answer 22

But I will leave squaring as soon as I get good in the other method : )

Answer 23

I don't really trust squaring

Answer 24

Me either

Answer 25

You get extraneous roots.

Answer 26

The other way splits it into four inequalities with potentially four possible solutions

Answer 27

Sometimes you only get two solutions, like in this case

Answer 28

I know but I need more practice at that.

Answer 29

squaring in this case is fine if you use equality

then pick numbers in your intervals to see if they satisfy the original equation

Answer 30

*original inequality

Answer 31

I need more practice at just inputting the correct equation. I used |3x+1| instead of |3x-1|

Answer 32

Thanks. @Zarkon

Answer 33

np