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Absolute Inequality fundamentals

Problem Solving — algebra and arithmetic (GMAT Focus Edition)
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by BarryLi » Sun Mar 13, 2011 12:56 pm
An inequality that involves one absolute value has 2 cases to consider.
Since this inequality has two absolute values, there are 3 cases to consider.
I'll discuss why there are 3 cases below.

Let a represent (3+2x)
Let b represent (4-x)

Now suppose |a| > |b| represents the original equation |3 + 2x| > |4 - x|

The 4 possibilities are as follows:

1) a > b
2) a > -b
3) -a > b
4) -a > -b

However, note that (4) * (-1) is a < b, which is the same inequality as (1) but with the inequality sign reversed.
i.e. (1) is a > b, (4) is a < b
The same is true for (3) * (-1), in that it is is the same inequality as (2) but with the inequality sign reversed.
i.e. (2) is a > -b, (3) is a < -b

Considering that, only inequalities (1) and (2) will need to be simplified to identify the 3 cases to consider.

1) a > b
3 + 2x > 4 - x
3x > 1
x > (1/3)

2) a > -b
3 + 2x > -(4 - x)
3 + 2x > x - 4
x > -7

From these two inequalities, the 3 cases to consider for the value of x are:

1) x < -7
2) -7 < x < (1/3)
3) x > (1/3)

Values for each of those ranges for x need to be plugged into the original inequality to see if they work.
If they keep the inequality true, then the numbers in that range can be a value for x.

1) x = -8 (for x < -7)
|3 + 2x| > |4 - x|
|3 + 2(-8)| > |4 - (-8)|
|3 - 16| > |4 + 8|
13 > 12 is true
x = -8 satisfies the inequality, therefore all values for x < -7 satisfy the inequality.

2) x = -3 (for -7 < x < (1/3))
|3 + 2x| > |4 - x|
|3 + 2(-3)| > |4 - (-3)|
|3 - 6| > |4 + 3|
3 > 7 doesn't hold
x = - 3 doesn't satisfy the inequality, therefore all values for x in the range -7 < x < (1/3) do not satisfy the inequality.

3) x = 1 (for x > 1/3)
|3 + 2x| > |4 - x|
|3 + 2(1)| > |4 - (1)|
|3 + 2| > |4 - 1|
5 > 3 is true
x = 1 satisfies the inequality, therefore all values for x > (1/3) satisfy the inequality.

Putting everything together, the values for x are:
1) x < - 7, and
2) x > (1/3)
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by shashank.ism » Mon Mar 14, 2011 7:10 am
gary391 wrote:What are the values of X?

| 3 + 2x | > |4 - x |
lets solve it this way..
case 1 : 3+2x>0 -->x>-3/2 and 4-x>0-->x<4
so | 3 + 2x | > |4 - x | --> 3+2x > 4-x --> 3x>1 --> x>1/3.
overall : 1/3<x<4.

case2 : 3+2x>0 --> x>-3/2 and 4-x<0 --> x>4
so | 3 + 2x | > |4 - x | --> 3+2x > x-4 --> x>-1.
overall : x>4.

case3: 3+2x<0 --> x<-3/2 and 4-x>0 --> x<4
so | 3 + 2x | > |4 - x | --> -3-2x > 4-x --> x<-7.
overall : x<-7.

case 4: 3+2x<0 --> x<-3/2 and 4-x<0 --> x>4
so | 3 + 2x | > |4 - x | --> -3-2x > x-4 --> x<1/3.
overall : No such solution.

at x = 4 also inequality is correct.

so the required value is [spoiler]x <-7 U x>1/3 . [/spoiler]
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by VivianKerr » Mon Mar 21, 2011 6:52 am
Barry Li's explanation is excellent. Just wanted to add a visual:

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