# The Holistic Approach to Absolute Values – Part V

We will continue our holistic approach to absolute values and add more complications to these types of questions. This article should set you up for any question of this kind. Note that this is a 750+ level concept, so if you are targeting a lower score, it may not be necessary for you to know.

(Before you continue reading, be sure to check out Part I, Part II, Part III and Part IV of this lesson.)

Let’s look at the following GMAT question:

For how many integer values of x, is |x – 6| > |3x + 6|?

(A) 1
(B) 3
(C) 5
(D) 7
(E) Infinite

In this question, we are given the inequality |x – 6| > 3*|x + 2|

Using the same logic as we did in the previous two posts, we will word the inequality like this: the distance from 6 should be more than three times the distance from -2.

At x = -2, the distance from 6 is 8 and the distance from -2 is 0. This means the distance from 6 is more than three times the distance from -2.

At x = -1, the distance from 6 is 7 and the distance from -2 is 1. Three times the distance from -2 is 3. This means the distance from 6 is more than three times the distance from -2.

At some point on the right of -1, the distance from 6 will be equal to three times the distance from -2. The distance between -2 and 6 is 8. If we split this 8 into 4 equal parts to get to x = 0, the distance from 6 will be equal to three times the distance from -2.

Now for every point to the right of 0, the distance from 6 will be less than three times the distance from -2.

Let’s try to go to the left of -2 instead. Will there be a point to the left of -2 where the distance from 6 will be equal to three times the distance from -2? Say that point is “a” units away from -2. -2 must then be 2a units away from 6 to ensure that 6 is a total of 3a units away from that point.

The distance between -2 and 6 is 8 – this 8 needs to be equal to 2a, so “a” must be 4 units.

The point where the distance from 6 will be equal to three times the distance from -2 will be 4 units to the left of -2, i.e. at -6. So at points to the right of -6 (but left of 0), the distance from 6 will be more than three times the distance from -2.

Note that for all values to the left of -6, the distance from 6 will be less than three times the distance from -2.

Hence, our x will lie in the range from -6 to 0.

-6 < x < 0

With these parameters, we will have 5 integer solutions: -5, -4, -3, -2 and -1. Hence, our answer is C.

Let’s look at a second question:

For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|?

(A) 1
(B) 3
(C) 5
(D) 7
(E) Infinite

Now the true value of this method is visible, as we have three or more terms. The arduous algebra involved in this given inequality makes our logical approach much more attractive.

First note that we have the term |5 – x|. This is the same as |x – 5| because |x| = |-x|.

We will word the inequality like this: the distance from 5 + the distance from 8 should be greater than the distance from -7.

Let’s find the point where the sum of the distance from 5 and the distance from 8 is equal to distance from -7. Say that point is “a” units to the left of 5.

a + a + 3 = 12 – a
a = 3

So the point is 3 units to the left of 5, which means it is at 2. For all points to the left of 2, the sum of the distance from 5 and the distance from 8 will be greater than the distance from -7.

How about the points that are to the right of 8? Say there is a point “b” units away from 8 where the sum of the distance from 5 and the distance from 8 is equal to the distance from -7.

3 + b + b = 15 + b
b = 12

So if we go 12 units to the right of 8, i.e. at x = 20, the sum of the distance from 5 and the distance from 8 is equal to the distance from -7.

For all points to the right of 20, the sum of the distance from 5 and the distance from 8 is greater than the distance from -7, so there will be infinite points for which the sum of the distance from 5 and the distance from 8 is greater than the distance from -7. Therefore, our answer is E.

Using this concept, try to answer the following question on your own: For how many integer values of x, is |x – 6| – |3x + 6| > 0?