Absolute Value Bugging me (1 Viewer)

[munkaii]

I haven't been taught to do the inequalities properly at school so i'm gonna rely on you guys. I've figured out a way to do the questions im about to post but im doubtful of possibilites of flaws in the method i use.

Here is the following: (Please state what region you are testing>>this is probably the part im most confused with)

|2x-1| - |1-x|<3

and

|2x-1| + |1-x|<1

I use the three case method.

 

[Slidey]

Three case method?

This method will always work.

1)
|2x-1| - |1-x|<3
(1) 2x-1+1-x<3
(2) 2x-1-1+x<3
(3) -2x+1+1-x<3
(4) -2x+1-1+x<3
x<3
3x<5
-3x<1
-x<3
.'. -3<x<3
Cases 2 and 4 were used.

2)
|2x-1| + |1-x|<1
(1) 2x-1+1-x<1
(2) 2x-1-1+x<1
(3) -2x+1+1-x<1
(4) -2x+1-1+x<1
x<1
3x<3
-x<1
-3x<-1
x>1/3
.'. 1/3<x<1
Cases 1 and 4 were used.

I see no pattern.

EDIT: Code clash. Used code tags to bypass active HTML detection
EDIT 2: God dammit. Maybe php will work.

 

[munkaii]

Three case method?

and

|2x-1| + |1-x|<1

1) |2x-1| - |1-x|<3
-2x+1+1-x<3
2x-1-1+x<3
2x-1+1-x<3
-2x+1-1+x<3
-3x<1
3x<5
x<3
-x<3
.'. -3<x<3

|2x-1| + |1-x|<1
2x-1+1-x<1
2x-1-1+x<1
-2x+1-1+x<1
-2x+1+1-x<1
x<1
3x<3
-x<1
-3x<-1
x>1/3
.'. 1/3<x<1

hey slide..did you just test all four cases then combine your answer?

 

[Slidey]

See my edit.

You should try and draw a graph. That way you know what you're looking for, just like with (x+3)(x-1)>0, for example.

 

[Trebla]

Use the graph method for the easy way out. However, you must be careful to use a very accurate scale. But chances are you'll get it wrong.
For: |2x - 1| - |1 - x| < 3
Graph: |2x - 1| < 3 + |1 - x|

For: |2x - 1| + |1 - x| < 1
Graph: |2x - 1| < 1 - |1 - x|

OR

You could use the 3 cases method. Much more accurate and effective than graphing but long:
For:
|2x - 1| - |1 - x| < 3
Critical points 1/2 and 1

x < 1/2
therefore |2x - 1| is negative and |1 - x| is positive
- 2x + 1 - 1 + x < 3
- x < 3
.: x > - 3
(combine that with condition x < 1/2 and find common range)
.: - 3 < x < 1/2

1/2 ≤ x ≤ 1
therefore |2x - 1| is postive and |1 - x| is positive
2x - 1 - 1 + x < 3
3x < 5
.: x < 5/3
(combine that with condition 1/2 ≤ x ≤ 1 and find common range)
.: 1/2 ≤ x ≤ 1

x > 1
therefore |2x - 1| is positive and |1 - x| is negative
2x - 1 + 1 - x < 3
.: x < 3
(combine that with condition x > 1 and find common range)
.: 1 < x < 3

COMBINE ALL SOLUTIONS
.: - 3 < x < 3

For:
|2x - 1| + |1 - x| < 1
Critical points are 1/2 and 1

x < 1/2
therefore |2x - 1| is negative and |1 - x| is positive
- 2x + 1 + 1 - x < 1
- 3x < - 1
.: x > 1/3
(combine that with condition x < 1/2 and find common range)
.: 1/3 < x < 1/2

1/2 ≤ x ≤ 1
therefore |2x - 1| is postive and |1 - x| is positive
*2x - 1 + 1 - x < 1
*.: x < 1
(combine that with condition 1/2 ≤ x ≤ 1 and find common range)
*.: 1/2 ≤ x < 1 (since range of answers in x < 1 does not include 1)

x > 1
therefore |2x - 1| is positive and |1 - x| is negative
2x - 1 - 1 + x < 1
3x < 3
.: x < 1
(combine that with condition x > 1 and find common range)
no common range .: no solution

COMBINE ALL ANSWERS:
.: 1/3 < x < 1

 

[munkaii]

You could use the 3 cases method. Much more accurate and effective than graphing but long:
For:
|2x - 1| - |1 - x| < 3
Critical points 1/2 and 1

x < 1/2
therefore |2x - 1| is negative and |1 - x| is positive
- 2x + 1 - 1 + x < 3
- x < 3
.: x > - 3
(combine that with condition x < 1/2 and find common range)
.: - 3 < x < 1/2

1/2 ≤ x ≤ 1
therefore |2x - 1| is postive and |1 - x| is positive
2x - 1 - 1 + x < 3
3x < 5
.: x < 5/3
(combine that with condition 1/2 ≤ x ≤ 1 and find common range)
.: 1/2 < x < 1

x > 1
therefore |2x - 1| is positive and |1 - x| is negative
2x - 1 + 1 - x < 3
.: x < 3
(combine that with condition x > 1 and find common range)
.: 1 < x < 3

COMBINE ALL SOLUTIONS
.: - 3 < x < 3

For:
|2x - 1| + |1 - x| < 1
Critical points are 1/2 and 1

x < 1/2
therefore |2x - 1| is negative and |1 - x| is positive
- 2x + 1 + 1 - x < 1
- 3x < - 1
.: x > 1/3
(combine that with condition x < 1/2 and find common range)
.: 1/3 < x < 1/2

1/2 ≤ x ≤ 1
therefore |2x - 1| is postive and |1 - x| is positive
2x - 1 + 1 - x < 3
.: x < 3
(combine that with condition 1/2 ≤ x ≤ 1 and find common range)
.: 1/2 < x < 1

x > 1
therefore |2x - 1| is positive and |1 - x| is negative
2x - 1 - 1 + x < 1
3x < 3
.: x < 1
(combine that with condition x > 1 and find common range)
no common range .: no solution

COMBINE ALL ANSWERS:
.: 1/3 < x < 1

Hey trebla.
For this region:
2.
1/2 ≤ x ≤ 1 you begin with equal to sign but u dont finish with one. Why is it the case?

1. So when you combine..
.: - 3 < x < 1/2
.: 1/2 < x < 1
.: 1 < x < 3

u get -3 < x < 3...

but where in combining have you showin that x=1/2 and1 is true..thats where i dont get it..

 

[Trebla]

Sorry my mistake, there should be an equal to as well for that section. I didn't check it and there's some errors in it. Check my edit. I put an asterisk on the parts I corrected.

 

[munkaii]

Edit: SOrry didnt realise you corrected it!

 

[littleboy]

AH ha ! i finally get it now. Thanks guys. No wonder i lost three marks for |2x-1| - |1-x|<3 in the first assessment task.