Absolute value

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ash4gmat: Senior | Next Rank: 100 Posts; Posts: 57; Joined: Thu Sep 17, 2015 2:57 am

by ash4gmat » Sun Dec 20, 2015 3:16 am

How to solve equations like these?

|7 - x| + |3 + x| = 10

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Sun Dec 20, 2015 3:46 am

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ash4gmat: Senior | Next Rank: 100 Posts; Posts: 57; Joined: Thu Sep 17, 2015 2:57 am

by ash4gmat » Sun Dec 20, 2015 4:08 am

Thanks Mitch, excellent explanation.

If equation would have been
|7-x| + |3-x| = 2

Then we have no solution.Right?

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Sun Dec 20, 2015 4:14 am

ash4gmat wrote:Thanks Mitch, excellent explanation.

If equation would have been
|7-x| + |3-x| = 2

Then we have no solution.Right?

Correct!

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ash4gmat: Senior | Next Rank: 100 Posts; Posts: 57; Joined: Thu Sep 17, 2015 2:57 am

by ash4gmat » Sun Dec 20, 2015 6:36 am

Mitch,

Can you also explain one example where we have difference of two absolute values.

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Tue Dec 22, 2015 6:08 am

ash4gmat wrote:Mitch,

Can you also explain one example where we have difference of two absolute values.

If |x+2| - |x-4| = 2, what is the value of x?

|a-b| = the distance between a and b.
|a+b| = |a - (-b)| = the distance between a and -b.

Question stem, rephrased:
|x - (-2)| - |x-4| = 2.

The first absolute value = the distance between x and -2.
The second absolute value = the distance between x and 4.

The greatest possible difference between the two distances is equal to the POSITIVE DIFFERENCE between the two numbers in the absolute values:
4 - (-2) = 6.
The least possible difference between the two distances is equal to the NEGATIVE DIFFERENCE between the two numbers in the absolute values:
-2 - 4 = -6.

The difference between the two distances will be equal to 0 when the two distances are EQUAL -- in other words, when x is HALFWAY between the two numbers in the absolute values.
Halfway between the two numbers = the average of the two numbers = (-2 + 4)/2 = 1.
If x=1, we get:
|x+2| - |x-4| = |1+2| - |1-4| = 3 - 3 = 0.
Here, the number line looks as follows:
-2<-----------x=1----------->4.

Every time x shifts one place toward the right endpoint (4), the first distance will INCREASE by 1, while the second distance will DECREASE by 1.
As a result, the difference between the two distances will INCREASE BY 2.
If x=2, we get:
|x+2| - |x-4| = |2+2| - |2-4| = 4-2 = 2.
If x=3, we get:
|x+2| - |x-4| = |3+2| - |3-4| = 5-1 = 4.
If x=4, we get:
|x+2| - |x-4| = |4+2| - |4-4| = 6-0 = 6.

If x>4, then the difference between the distances will remain 6, since 6 is the maximum possible difference.
If x=5, we get:
|x+2| - |x-4| = |5+2| - |5-4| = 7-1 = 6.

Every time x shifts one place toward the left endpoint (-2), the first distance will DECREASE by 1, while the second distance will INCREASE by 1.
As a result, the difference between the two distances will DECREASE BY 2.
If x=0, we get:
|x+2| - |x-4| = |0+2| - |0-4| = 2-4 = -2.
If x=-1, we get:
|x+2| - |x-4| = |-1+2| - |-1-4| = 1-5 = -4.
If x=-2, we get:
|x+2| - |x-4| = |-2+2| - |-2-4| = 0-6 = -6.

If x<-2, then the difference between the distances will remain -6, since -6 is the least possible difference.
If x=-3, we get:
|x+2| - |x-4| = |-3+2| - |-3-4| = 1-7 = -6.

As illustrated by the portion in red, |x+2| - |x-4| = 2 when x=2.

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