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# Absolute Values For GMAT/GRE

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#### Josh Jones

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Absolute values come up very often in official exams, and are typically combined with equalities or inequalities. They can typically represent two equations or inequalities in one line. Here I will introduce a method to make solving absolute value functions less intensive, and a few tricks to make absolute values easier than you think!

| 3x – 5 | = 10
For this equation, think about what 3x – 5 could be to satisfy the equality. Since it’s inside an absolute value, we could have 3x – 5 = 10 and let | 10 | = 10. Or, find the negative counterpart and see that | -10 | is also equal to 10. Therefore, 3x – 5 must equal to 10 or -10, and these are two solutions for x. Note that there is a neat way to represent this relation without absolute values. We square both sides to get:
(3x – 5) ^ 2 = 100
Then there is no absolute value sign in the equation, and to solve this equation we would prefer square rooting both sides to get the absolute value equation above. Note that square rooting both sides is only acceptable when both sides are guaranteed to be positive or 0. With this representation, we clearly have quadratic terms and therefore should have two solutions. This can help remind you that typical absolute value equations should also have two solutions.

Now, let’s focus on an absolute value inequality.:
| 3x – 5 | < 10
Here we have the same equation as above, but this time with inequalities. To solve this, we again think about what 3x – 5 needs to be. We want 3x – 5 to be less than 10, to make | 3x – 5 | not exceed 10. But then we also want it to not be smaller than -10, otherwise the absolute value of that number is greater than 10. So we have the following equations:
3x – 5 < 10
3x – 5 > -10
If we graph this on a number line, we get the following:

Notice how both statements are combined with an “AND” conjunction. Both need to be satisfied simultaneously. Allow me to introduce a trick to help memorize these equations. First, transpose the second equation and then line them up.:
3x – 5 < 10
-10 < 3x – 5
Any observations about these two equations now? If you found something interesting, then you are a great mathematician! We can combine these two inequalities into one double inequality.:
-10 < 3x – 5 < 10
Now this just means, from | 3x – 5 | < 10 we know 3x – 5 must lie between -10 and 10. The convenience here is that we can proceed to solve normally but we can solve two equations at once. Typically, we would add 5 to both sides, then divide both sides by 3 to get x. Here we add 5 everywhere and divide by 3 everywhere.:
-10 < 3x – 5 < 10
-5 < 3x < 15
-5/3 < x < 5

We just solved two equations at once! In just two to three steps, this technique is very useful when we have | xxx | < #, but it could be used when the inequality sign goes the other way, as well. Let’s say the inequality is:
| 3x – 5 | > 10
The normal solution is to let 3x – 5 > 10 or 3x – 5 < -10. (Note that this time it’s an “OR” conjunction since we can have either case). We can try combining anyway and see what happens.:
3x – 5 > 10  OR -10 > 3x – 5
– 10 > 3x – 5 > 10
– 5 > 3x > 15
-5/3 > x > 5
Apparently, this doesn’t make sense. But let’s just split the inequality and take the results, that x > 5 or x < -5/3 is still a correct answer. We need to split up the inequality at the end and interpret the numbers correctly. Let’s also take a look at the graph representation.:

This shaded region is just the part that was not shaded from the graph above. Therefore, this inequality completes the other inequality with the opposite sign, which makes sense since they are opposite regions.

Here are some other minor tips.:
| -2a + 10 | < 16
is the same as
| 2a – 10 | < 16
We can flip the sign inside the absolute value since it doesn’t matter if it’s positive or negative inside. It’s the same result after absolute valuing it.
| 2a – 10 | < 16
is the same as
2 | a – 5 | < 16
| a – 5 | < 8
We can treat the absolute values as brackets when you want to take out factors. We can take out any factor but it must be positive. If we are not sure if it is positive, we can force the factor to be positive with absolute values, hence:
| x*a – x*b | = | x * (a – b) | = | x | * | a – b | 