GMAT Math will ask you about absolute values. Mastering what the GMAT asks about them requires sophisticated understanding.

Somewhere along the line, perhaps in middle school, you probably learned:

|positive| = positive and |negative| = positive

In other words, the equation |x| = 5 has the solution: x = 5 **or** x = −5. (Notice: the word “or” is not a garnish there; it’s actually an essential piece of mathematical equipment.)

## Expanding the Pattern

That’s great, but the GMAT is simply not going to ask you to solve the equation |x| = 5. When the GMAT asks about absolute value, it’s going to be something more in the vein of |3x – 7| = 5. The basic idea is (as is often the case in more advanced algebra) is to replace “x” in the simpler equation above with whatever “thing” is between the absolute value. If q is a positive constant, then

|thing| = q

has the solution:

thing = q **or** thing = -q

In the given example,

|3x – 7| = 5

3x – 7 = 5 **or **3x – 7 = -5

3x = 12 **or** 3x = 2

x = 4 **or** x = 2/3

That’s an example of an absolute value equation, which the GMAT could ask. The GMAT is even more likely to ask about an absolute value inequality.

## Rethinking Absolute Value

OK, let’s face it. The definition of absolute value that says “keeps a positive positive, and makes a negative positive” – the utility of that definition peaks in middle school. We need to have a more sophisticated understanding of absolute value to handle everything the GMAT will ask of it.

Here is the more sophisticated definition of absolute value. The absolute value of x, |x| is **the distance of x from zero on the number line**. Of course, it’s always positive, because distance is away positive.

To extend that further: |x – 4| is the distance of x from 4; |x – 7| is the distance of x from 7;|x + 3| is the distance of x from -3 (this is because x + 3 = x – (-3) when written as subtraction).

That is profoundly important in solving the absolute value inequalities that the GMAT will ask of you. Suppose a GMAT Math question asks you: represent the region -1 < x < 9 as an absolute value inequality.

The first step is to find the midpoint of the region: 4 is exactly halfway between -1 and 9. Now the distances: 9 is a distance of 5 from 4, and so is -1. So the distance from 4 (viz. |x – 4|) can’t equal 5, but it can be anything up to 5. Thus

|x – 4|< 5

is the absolute value inequality representation of the region -1 < x < 9. Integrate this understanding, and you will be able to handle anything the GMAT asks you about absolute value.

Here’s a free Magoosh practice problem, a challenging absolute value practice question, with a video explanation of the answer.

### Most Popular Resources

Actually, I just thought through this some more. Wouldn’t it be |x-4|<5?

Thanks.

Hi Mike,

Great article. But I am not clear on how u explained the least question by Manisha.

I understood ‘y’ is negative. But when y is negative |-y| becomes y and -(-y) becomes +y.

And so the equation should be x+2y=6. Isn’t it? Please help me!!

Thanks,

John.

if y = negative number, |y| = -y

lets do this with example…

say y = -1

|y| = |-1| = 1

-y = -(-1) = 1

so, if y = negative number then |y| = -y

Hi Mike! Thank you for updating our understanding of absolute values. In the spirit of “playing” with the numbers as your video suggests, how would the solution change if we were asked to represent the region when one of the signs is greater than/equal to? for example, -1 is less than x and x is less than OR equal to 9. How could we show that in the resulting absolute value inequality?

Thanks!

Cassie,

Great question! My friend, the short answer is: we can’t. You see, the absolute value, by its very nature, is a deeply symmetrical entity. We can’t use something symmetrical to represent something asymmetrical. If we have the “or equal too” tag at both endpoints, or at neither endpoint, we can use the absolute value. An segment with one closed endpoint and one open endpoint is a profound asymmetrical item, and absolute value can no longer help us.

Does this make sense?

Mike 🙂

Is there any way the inequality can be a positive? like, abs(x+4)<5? How would this look like? I guess I have the same problem as Anush where I don't see why it's a negative, not a positive. Yes I can plug in numbers but the concept itself doesn't make sense to me.

Thanks!

Dear Kay,

I apologize for a belated reply. Yes, instead of subtraction, we could have addition in the absolute value. Remember that (x + 4) = [x – (-4)], so it’s the distance from x = -4. Think about it. At x = -4, |x + 4| equals zero. At x = -3 and x = -5, it equals 1. At x = -2 and x = -6, it equals 2, and so forth. Thus, the inequality |x + 4| < 5 would equal everything that was less than a distance of 5 units from x = -4, so it would be -9 < x < +1.

Again, as I urged Anush, I will urge you to draw a number line on paper and

seethese distances. This is primarily a visual concept, and you will not get it purely with words. You have toseeit.Does all this make sense?

Mike 🙂

Thank you for the article! Very helpful.

And what if we had the following equation?

|1-x|>4?

Does it mean that the distance between x and 1 is greater than 4?

Dear Yulia,

You are more than welcome, my friend. 🙂 I’m very glad you found it helpful! As to your question, remember that when we reverse the order of subtraction, the only difference is a negative sign: (x – 1) = -(1 – x). When we take an absolute value, that negative sign goes away: |x – 1| = |1 – x|. The fact that there is an equal sign means that those two are unutterably identical. There is absolutely no difference at all between |x – 1| and |1 – x| — if either is greater than four, then the distance between x & 1 is greater than 4.

Does all this make sense?

Mike 🙂

Hi Mike,

This is probably a dumb questions but can you tell me why it’s x-4 not x+4?

Thank you.

Mara

Dear Mara,

That’s certainly not a dumb question! Remember, my friend: every letter in algebra stands for a number, so in order to understand the algebra, it always is helpful to try out a few sample numbers. The question is: if we pick a new point on the number line, how far is this new point from the point x = 4? Think about if we picked x = 3 as our new point: what mathematical operation would we perform get the distance between 3 and 4? We would subtract. Again, suppose our new point were x = 6: once again, we would subtract to find the distance. That’s why it’s (x – 4), not (x + 4). The minute something in algebra doesn’t make sense, plug in numbers to check out how numbers would behave in the same situation. Does all this make sense?

Mike 🙂

Hi Mike,

Yes it does. Sometimes I overthink it and it’s just a simple answer! Thank you!

Dear Mara,

You are quite welcome! 🙂 Best of luck to you in all your studies.

Mike 🙂

This is a great article on absolute values Mike, I never thought of absolute values like this

🙂

Dear Johann,

You are quite welcome! 🙂 I am very glad you found it helpful! Best of luck to you!

Mike 🙂

I think there is a minor typo….

|3x – 7| = 5

3x – 7 = 5 or 3x – 7 = -5

3x = 12 or 3x = 2

x = 4 or x = 3/2

I think that last x should be 2/3, vice 3/2.

Other than that, thanks for the blog! I’ve learned a ton in the last few days, thanks to you, Mike.

-Shawn

Dear Shawn,

Thank you for pointing that out! 🙂 Great eye! I just updated that. And thank you for you kind words! I am very glad you have found this helpful. Best of luck to you!

Mike 🙂

Hi..I don’t understand the way you found the region -1 < x < 9 as an absolute value inequality. I understood that midpoint is 4 , and 9 is distance of 5 from 4..then what about -1.

and the distance from 4 (viz. |x – 4|) can’t equal 5, but it can be anything up to 5. Thus

|x – 4|< 5?? I am lacking some basic things:(.

Pls explain it:)

Dear Anusha,

I’m happy to respond. 🙂 Imagine it this way. Imagine standing on the number line, at x = 4, the midpoint. If I walk 5 units to the right, I will be at +9. If I walk 5 units to the left, I will be at -1. I urge you to draw a number line on paper and verify this for yourself. The points -1 & 9 are the two points at a distance of 5 from the point 4. The allowable points have to have

lessdistance, so they arebetween-1 and 9. Again, I urge to you draw this out on paper so you can see it.Does all this make sense?

Mike 🙂

Hi… The solution to the “a challenging absolute value practice question” says that y must be negative but in that case mode of y would be +ve y and -y would also be +ve, therefore resultant eqtn should be x+2y =6 and not x-2y=6

Manisha: This is subtle point that confuses many people. Technically, the “-” sign as three different roles:

(a) in front of a number, it’s a minus sign

(b) between two terms, it’s a subtraction sign

(c) in front of a variable, it’s an OPPOSITE sign.

In particular, if we know that y must be a negative number, as we deduce in this solution. the y is negative, and (-y), using the opposite sign, is positive, because the opposite sign changes a variable to the opposite of its original value. Thus, if we know y is negative, we know |y| = -y —- again, in front of a variable, that is NOT a negative sign, but rather an *opposite* sign. Thus, the equations that appear in the solution are correct.

Does all this make sense?

Mike 🙂