On average you will see at least one question on the Revised GRE dealing with absolute values. You may even see a few. Yet, absolute value gets lost in the prep fray amongst the more popular concepts. So if you don’t want this relatively innocuous concept to surprise you test day read on.

## What You Need to Know

What do -4 and 4 have in common? They are both four units from the 0 on a number line. Think of absolute value has how far from the zero a given number is on a number line.

For positive numbers finding the absolute value is easy – it is always the number between the absolute value signs, which look like this .

When we take the absolute value of a negative number, we drop the negative, and the absolute value sign.

## Absolute Value Meets Algebra

What is the value of x in the equation ? Well, the absolute value of both and

Now take a look at the following:

Anything seem off? Well, the absolute value of any number can never be a negative therefore there is no value for x.

To solve for a variable inside an absolute value sign, we want to remove the absolute value sign and solve the equation. However, there is a slight twist: you will want to create two separate equations. For one remove the absolute value signs and solves for x. For the other, make the side of the equation not inside the absolute value equal to a negative. Let’s try a practice problem:

Our two equations are:

If this seems strange, think of it this way: when you find a value for x that makes the equation equal -4, you have also solved for positive four. Remember, the absolute value of -4 is 4. Solving for x we get:

Therefore x can equal 6 or -2.

## Absolute Value Meets the Inequality Sign

Now let’s complicate things a little and throw an inequality in to the picture. Have a look:

We turn the inequality sign into an equal sign and solve for x the way we did above.

so

However, when we turn 3 into -3, we have to reverse the sign. This is the one critical step.

so

## Takeaways

This is a basic overview to absolute value and should help you with most of the sub-150 problems. For the harder problems, however, you will want to make sure to practice with more advanced problems.

|x-5| -> Distance of x from +5

This is not clear, can you explain the concept behind it? Why cant it be distance from -5.

|x-5| really does represent the distance of x from +5, and |x+5| really is the distance of x from -5. This can seem confusing at first, Navaneeth, but it makes sense if you plug in values for x. Let’s give that a try.

Suppose that x=8. Then, |x-5| becomes |8-5|. This simplifies to |3|, or just 3 (since absolute values are positive numbers anyway). If x = 8, in other words, x is 3 units from 5, from positive 5. It is NOT 3 units away from -5; 8 is actually 13 units away from negative 5. If we set x at 1, we still get the same result of x being the distance from positive 5 rather than negative 5:

|x-5| >> |1-5| >> |-4| >> 4. 1 is 4 units from 5. (But 6 units away from -5.)

This really will work with any value for x: a value greater than 5, a value less than 5, a negative number, or a positive number. If you still have some doubts about the number properties demonstrated above, practice plugging other values in for yourself.

The same principle is the reason |x+5| is the distance of x from negative 5. Let’s plug in two different random numbers for X. First, let’s say that x= -100. Here’s how that plays out:

|x+5| >> |-100+5| >> |-95| >> 95. -100 is 95 units away from -5. (But would be 105 units away from positive 5.)

Another way to think of it is this: |x-5| means that |x-5| is a certain number greater than, less than, or equal to 0. It’s an unknown equality, expressed as: |x-5| ? 0. Recall that inequalities can be treated like regular algebraic equations. So to get x alone on one side of the equation, you want to eliminate -5. You do that by adding +5 to both sides: |x-5+5|?0+5 >> |x|?5, with “?” representing not just inequality to 5, but specific distance in either direction from 5. Similarly:

|x+5|?0 >> |x+5-5|?0-5 >> |x|?-5 .

Hope all this helps. 🙂

I have a question, the lesson shows an example

| x – 1 |> 4, the results are x> 5 and x ,<) changed. Because the rules to change the sign only apply to division and multiplication with negative numbers.

Hi Linda,

Since you’re a Premium student, I forwarded your message on to our team of tutors. You should hear back from them soon! As a reminder, you can always get answers more efficiently by using the Purple “Help” button in your Premium account 🙂

Hey! The sign has to be changed because you’re actually multiplying “-1” to the expression in the right-hand side. For example:

|x-3| > 3, when you take out the absolute value, it becomes:

x-3 > 3 and x-3 < -3.

Hope this helps! 🙂

Chris,

I am a Clemmonsdogparker and I have a doubt here. If we have an equation like |x-5| + 3 = 8. then how to solve this for the range of x.

Following your method of solving the equation, i reached the conclusion x= 10 or -6 however the answer chances given as follows in the O’Level math book are as follows.

|x-5| + 3 = 8

A. x=10

B. x= 10 or x=0

C. x=10 or x=-10

D. x=0 or x=-10

I solved as follows:

|x-5| + 3 = 8 or |x-5| + 3 = 8

x-5+3=8 or x-5+3=8

x =10 or x =-6

The correct answer as per the book is B.

Can you please explain how we reached this result.

Hi Muqadas,

If you get a problem like this, you want to try to make your life easier and get the absolute value onto one side of the equation by itself:

|x-5| + 3 = 8 [Subtract “3” from both sides]

|x-5| = 5

Then, we have:

“|x-5| = 5” OR “|x-5| = -5”

|x-5| = 5 [Remove absolute values sign (as we have both equations now)]

x-5 = 5 [Add “5” to both sides]

x = 10

|x-5| = -5 [Remove absolute values sign (as we have both equations now)]

x-5 = -5 [Add “5” to both sides]

x = 0

So, I agree that the correct answer is B.

Thanks Chris for the prompt reply.

Kindly take a look at another example question 5-3 |x+5| = -4 which I am currently trying to solve by the above-mentioned method to results that aren’t matching any of choice.

The answer choices are as follows:

A. x= 4 or x= 10

B. x= -4 or x= -10

C. x= 4or x= -4

D. x= No Answer Possible

I solved it as follows:

5-3 |x+5| = -4

2 |x+5| = -4 Subtracted -2 from both sides – |x+5| = -6

x+5 =-6 or x+5 = +6

The answers were -13 and -1 but the true answer option by the book is B.

Kindly tell me how was this result drawn.

Hi Muqadas,

I’m a bit confused by this example–I think you may have copied something incorrectly in the question itself. I didn’t get an answer included in the choices.

The problem I see with your answer is that you added and subtracted unlike terms. From what I see, the 3 is multiplied by |x+5|, which means that we can’t subtract 3*|x+5| from 5 (this would be like doing 5-3x=2x– there is no x in the 5 so they aren’t like terms and we can’t do it!)

This is how I solved the problem using the method above:

5-3*|x+5|=-4

-3*|x+5|=-9

|x+5|=3

So we have |x+5|=3 OR |x+5|=-3

For the first equation:

x+5=3

x=-2

For the second equation:

x+5=-3

x=-8

So I’m not sure if the question was copied down incorrectly or not, but I didn’t get any of the answer choices here! I recommend that you double check the problem to see what might be going on, and see if my method makes sense to you. Remember: you can’t add or subtract different terms! 🙂

I’m a bit confused by the above example and how the answers are not listed.

I’m wondering: does this concept, “the absolute value of any number can never be a negative therefore there is no value for x” not apply in this situation because carrying out/simplifying the operations on the left side produces a positive number on the right side once executed?

Hi Adrienne,

So, I think I understand where your confusion is coming from here! So, please note that you cannot have an absolute value expression

aloneon one side of the equation EQUAL TO a negative numberaloneon the other side. For example, |x + 5| = -8. This does NOT work. As you’ve pointed out, an absolute value cannot equal a negative number as the absolute value expression will output a positive number.That being said, if there is a negative expression with the absolute expression on one side of the equation, then it’s possible for the other side of the equation to be negative. For example, |x + 5| – 20 = -8. In this case, this is possible. But, you have to simplify to confirm.

I hope this helps! Have a great day! 🙂

Hi Chris,

How can we solve the Quantitative comparison for absolute value solution and 0?

Eg: |2n+1| < n +6

Quantity A: n

Quantity B: 0

Can you please help.

This problem is in Official Guide.

Hi Ketki,

In this case, I would recommend plugging values in for n to see what the relationship might be. This has an absolute value and an inequality, so it’s a difficult question! But, if we can determine that there are values that satisfy n that are greater than and/or less than 0, we can figure out relationship out. I can see pretty quickly that I can plug “2” in for n and get 5<10, so this inequality holds true for a value greater than 0. I can also see that if I plug in -1 for n, I get 1<6 which also holds true. Since there are values above and below 0 that satisfy this inequality, we know that the answer must be (D).

For more information on the plugging in strategy, see the following blog section: http://magoosh.com/gre/category/math-strategies/plugging-in/

Hi,

I came across this question. Can you please answer it?

Suppose that |x|<|y+2|0 and xz>0. Which of the following could be true?

Indicate all statements that apply.

a) 0<y<x<z

b) 0<x<y<z

c) x<z<0<y

d) 0<y+1.5<x<z

e) z<x<0<y

Hi Sakshi,

I’d be happy to help you with this question, but could you let me know where it’s from? Please note that we generally don’t answer questions from outside materials but rather prioritize Clemmonsdogpark and Official Guide questions 🙂

Also, it’s a great help to us in giving explanations if you elaborate a little about your thought processes. How did you approach the question? Where did you get stuck? This helps us to make clear, concise responses.

Looking forward to hearing from you!

Hey ,

I just want to clear my doubt .

The absolute value of a number is always positive , whereas when the variable is there it can be negative or positive . Right?

|4| = 4 , |-4| = 4

|x-4| = (x-4) or -(x-4)

Hi Sayli 🙂

Happy to help! You’re headed in the right direction with your thinking, but let’s tighten it up 🙂

|something| is the distance of (something) from zero. It’s a distance, which means it’s always positive.

As you mentioned, |4| = |-4| = 4. And another way of saying this is that both 4 and -4 are a distance of 4 units from 0.

VariablesNow, let’s look at what happens when we have a variable inside the absolute value sign. For example,

|x| = |-x|

But do these equal x or -x? Well, it depends on whether x itself is negative or positive. Remember, -x doesn’t mean a negative number necessarily. It means the “opposite of x.”

So if x = -3, then -x = -(-3) = 3

If x is a negative number, then -x is a positive number. So if x is negative, then |x| = -x. This idea reflects the example you mentioned 🙂

Hope this helps!

This is the problem |3 + 3x| < -2x

Quantity A Quantity B

|x| 4

so, x < -3/5 <<>> x>-3 ( 5>-3 is true here )

how can we say => -3/5 > x > -3 (This means x < -3/5 <<>> x>-3)

and conclude that Quantity B is greater

Hi John,

Happy to help 🙂 When we solve this absolute value inequality, we get find, as you mention,

-3 < x < -3/5

In words, x is greater than -3 and less than -3/5. So, there is a range of values from -3 to -3/5 (these endpoints not included) that x could be and satisfy the original inequality. Given this range, |x| < 3, since x < -3 and |-3| = 3. And for that reason, we can conclude the Quantity B is greater 🙂

I hope this helps!

I thought you only switch the sign when multiplying/dividing by a negative, not adding/subtracting?

Hi Tee 🙂

You’re correct that switch the sign of an inequality when we multiply or divide by a negative number and we keep the sign the same when adding or subtracting. We can see these ideas applied in the practice problem in the post 🙂

|x-4| < 3

To solve for x, we want to look at the positive and negative of the value within the absolute value:

1. x – 4 < 3

2. -(x – 4) < 3

As outlined in the post, for (1), we solve for x by adding 4 to both sides:

x – 4

+ 4< 3+ 4x < 7

For (2), we can first multiply both sides by (-1), which entails flipping the sign of the inequality. From there, we will add 4 to both sides to solve for x:

-(x – 4)

*(-1)< 3*(-1)x – 4

>-3 [notice how we flipped the sign of the inequality!]x – 4

+ 4> – 3+ 4x > 1

Combining these two results, we find that 1 < x < 7 🙂

I hope this clears up any confusion! Happy studying!

Hi Chris,

I am a Clemmonsdogparker. Could you please delineate the concept of extraneous roots?As it was mentioned by Mike that ‘extraneous solutions are invalid and do not solve the original equation’ in a lesson video!!

Hi Sayahnita,

Happy to explain! 🙂

An extraneous root is one that appears as a solution, but upon checking with direct substitution is in fact not a solution.

For example: you have an equation and after some work come up with two roots which we can juts call “a” and “b”. When you put “a” into the original equation it works and is true, but when you put in “b” it doesn’t work at all! So “b” is an extraneous root because while you arrived at it in a mathematically valid way, it isn’t actually a root. This often happens when we square both sides during our solution because, as you know, this affects negative numbers. (The square of a negative number is positive.)

Imagine we had to solve √(2x + 7) + 4 = x.

√(2x + 7) = x – 4

2x + 7 = (x – 4)^2

2x + 7 = x^2 – 8x + 16

x^2 – 10x + 9 = 0

We can factor this, which gives us (x – 9)(x – 1) = 0. Thus, 9 and 1 are roots, but we now have to go back and check them in the original equation. If you plug x = 1 back into the original equation, you will see that the equation DOES NOT hold true. Hence, this is an extraneous root. If you plug x = 9 back into the original equation, you will see that the equation DOES hold true, so that one is your answer.

For 1:

√(2x + 7) + 4 = x

√(2(1) + 7) + 4 = 1

√(9) + 4 = 1

3 + 4 =/= 1

For 9:

√(2x + 7) + 4 = x

√(2(9) + 7) + 4 = 9

√(18 + 7) + 4 = 9

√(25) + 4 = 9

5 + 4 = 9

I hope that helps!

Wow,gee thanks,you guys are the best 🙂

I have a question about absolute value equation,

if we get two negative solutions for X, so we directly can say there is no solution for absolute value equation

Why we need to solve in two Equations a, + 3 B, -3 to find the absolute values? Is it to know the number of units?How about the Below one?

Quantity A: l m+25 l

Quantity B: 25 – m

Thank You.

expecting answer from Chris

hello chris i am a magoosher, could you please explain the following ; |-R/4+6|>2 and

|-R/4+6|>-2

Hi Pranesh!

Thanks for pointing out that you’re a Clemmonsdogparker! 🙂

Just wanted to let you know that I’ve forwarded your question to our team of Remote Tutors. Chris would love to answer every question, but he has a lot of projects going on right now and he’s just one person! Someone from the tutor team will email you directly to follow up.

All best,

Jessica

Hi Chris

I am a Clemmonsdogpark premium member. I really like your blogs and videos. Great work!

One thing I did not understand in the above blog was its last part. When solving x-4=-3, why did you reverse the inequality sign? Clearly the equation does not look like its getting multiplied or divided by -1, since that would have affected both the sides.

Mudit,

When we make the number to the right of the equal sign a negative number (in this case -3), it reverses the direction of the sign. Take the following equation:

|x + 1| < 2

One value of 'x' has to less than 1 (we can solve this the traditional way).

Now we have to find the value of x that would make the equation x + 1, now without the absolute value sign, come out to less than -2. See, whatever value that comes out of the absolute value equation that is between 0 and -2 will yield a positive number (remember the absolute value sign). For that reason we set the number equal to -2. But in doing so, we have to flip the direction of the sign. Remember when we solved for the positive value of x, we get a number less than 2. To say that x has be a number less than -2 would not overlap (these statements are mutually exclusive). By turning the direction of the sign, when we make the number -2, we make it so 'x' is between -2 and 2, which is a possible outcome.

Hope that helps!

its simple that why reverse the inequality sign . for example

2>1

now multiply by -1 on both sides

-2>-1 is wrong

so we reverse the sign

-2<-1

now do u get it

Hi Chris,

Can you please share me the magoosh link for the chapter “sequences”. I tried in Algebra and Math Basic sections but didn’t get.

Thanks,

Vikas

Hi Vikas,

Are you talking about the eBook?

In you above explanation to the question: What is the value of x in the equation delim{|}{x}{|}=5. Why there is no value of x? Can’t x be -5 and 5 in this?

Is it alright to apply these strategies with quadratic inequalities?

Where can i find some harder problems on absolute values

Hi Peter,

The Clemmonsdogpark products offers some, the Manhattan 5 lbs. GRE book is also a great source, and the NOVA book (old GRE only) may have a few.

Hope that helps!

Chris,

I have a doubt here. If we have a equation like |x-1| + |x-2| = 5. then how to solve this for range of x.

Hi Shubham,

All you have to do is remove the absolute value signs and solve for x, giving us x = 4. Then, remove the absolute value signs but this time make 5 negative, giving us x = -1.

Hope that helps!

hi! in the previous question,can i know why 5 is taken as -5?

Yes, whenever you are dealing with absolute value signs, you always want to make the number to the right of the equal sign negative to solve for the second value of the absolute value.

|x – 4| = 1

To solve, we remove the absolute value signs and create two equations:

x – 4 = 1

x – 4 = -1

Then solve for x to find the two values of ‘x’.

Hope that helps!

In the above explanations when we do have two absolute values, wont the negative and negative become positive and we should only consider the positive on the right hand side? Or is it always that you consider the RHS to be -ve and +ve both irrespective of the number of absolute values on the LHS.

Thank you- you simplify this!!

Really helpful Chris, Thanks a lot! 😀

I think it’s not so straight to solve |x – 1| + |x – 2| = 5 by just reversing signs. The approach doesn’t seem to be extendible to an example like |x – 1| + |x – 2| + |x – 3|+ |x – 4| .. = 5

The steps should be:

1. Split into parts

2. Check sign in each part & solve the equation

3. Find solution and then validate it

Explanation: