# Lesson: Absolute Value Equations

## Comment on Absolute Value Equations

### Hi, Just a quick question.

Hi, Just a quick question. When you say to plug our possible solutions to the original equation, is the
-(negative) equation not considered part of the original equation? I ask because i'm confused. If{x}=3x-4 or -3x-4;
my first solution of 2 i plugged into 3x-4. my second solution of 1 i plugged into -3x-4 because I thought the solution of 1 was derived from that "original equation of -3x-4. So for future reference, when I am tackling absolute value questions, I am always plugging the solutions into the first equation only?
thanks

### You must plug the solutions

You must plug the solutions into the ORIGINAL equation (not the derived equation). Otherwise, you are just confirming what you already solved.

That is, from the equation |x| = 3x - 4, you derived two equations: x = 3x - 4, and x = -(3x - 4)

When you solved x = 3x - 4, and got x = 2, you got that solution by solving the DERIVED equation (x = 3x - 4), so it doesn't help to then plug that solution into the (derived) equation that you just used to determine that solution.

If you follow that strategy, you will never find any extraneous roots.

Our goal is to solve the original equation, so we must use that equation to check our possible solutions.

### Hi Brent,

Hi Brent,
https://gmatclub.com/forum/what-is-the-sum-of-all-the-real-values-of-x-for-which-x-4-2-x-242038.html
You used what you called U-Substitution which is great idea to simplify the original equation, but for someone like me and many non-math experts, this idea will not always come in mind.
So I tried to simplify the equation as follows:
|x-4| * |x-4| + |x-4| = 30
x^2 -4x -4x +16 + x-4 = 30
x^ -7x -18 =0
(x- 9) (x+2) = 0
x=9 or x=-2

Just wondering if the way I am using to simplify the equation that includes absolute value is valid approach?

Thanks.

Unfortunately, that approach will tend to miss potential solutions (as it did here).

Also, as with all equations involving absolute value, you must always confirm whether any of your solutions are extraneous roots.

When we test your solution x = -2, we find that it is NOT a solution to the original equation.

Cheers,
Brent

### Hi Brent,

Hi Brent,
https://gmatclub.com/forum/if-x-4-what-is-the-range-of-the-solutions-of-the-equation-14-x-239196.html
I have used the method of plugging in the answer choices and I got 8 is the only number that was valid. I found it faster than the algebraic method in this case.
I am not sure if that was a good approach in this kind of questions or it was coincident.
Thanks

Your solution is a fortunate mistake. Yes, it yielded the correct answer, but that was just a coincidence.

I say this because you answered a question that's different from the question that is asked. The question asks us to find the RANGE of all possible solutions.

When we solve the given equation, we find out that there are 3 solutions: x = 8, x = 10 and x = 16

So, the RANGE = biggest value - smallest value
= 16 - 8
= 8

So, when you plugged in the answer choices, you found that x = 8 is ONE solution. This alone is not enough to determine the RANGE of all possible solutions.

However, since you misread the question as "What is the solution of the equation |14–x| = 24/(x−4)?", you entered answer choice C (which turns out to be the correct answer .... to the real question).

Does that help?

Cheers,
Brent

Absolutely,
Thanks Brent.

### hi Brent - thanks for posting

hi Brent - thanks for posting this question (https://gmatclub.com/forum/if-x-and-y-are-positive-integers-what-is-the-value-of-x-238519.html#p2024805) and providing the explanation. when working with absolute values, I thought we would have to consider all possibilities as follows - please help me understand why this approach is incorrect (and since I ended up with 4 possibilities as shown below - I concluded that this statement was insufficient). I can see why case 2 and case 3 are invalid as RHS and LHS are not equal - but what about cases 1 and 4?:

case 1: 3x + 5y = 21

case 2: -(3x + 5y) = 21

and since we took the square root of both sides we were left with + / - 21 on the RHS, so we have two more cases:

case 3: 3x + 5y = -21

case 4: -(3x + 5y) = -21

Good question.

You'll find that the solution to the case 1 equation is identical to the case 4 equation, and the solution to the case 2 equation is identical to the case 3 equation. The reason for this is that equations 1 and 4 are equivalent equations, and equations 2 and 3 are equivalent equations.

Take, for example, case 1: 3x + 5y = 21
If we multiply both sides of this equation by -1, we get -(3x + 5y) = -21 (the case 4 equation).

The same holds true for equations 2 and 3

Does that help?

Cheers,
Brent

How come in your solution we can ignore the absolute value in the RHS? Rephrased: do we not need to account for the fact that |x+1|/2 can be positive or negative?

Also, is there a way to solve both sides together with squaring, and why would that work mathematically?

Thanks!

In my solution (https://gmatclub.com/forum/if-x-1-2-x-1-what-is-the-sum-of-the-roots-244...), I cover all possible cases.

GIVEN: |x+1| = 2|x-1|

So, EITHER x+1 = 2(x-1) OR x+1 = -[2(x-1)]

Notice that the equation x+1 = -[2(x-1)] is equivalent to the equation -(x+1) = 2(x-1)

Likewise, the equation x+1 = -[2(x-1)] is equivalent to the equation (x+1)/2 = -(x-1)

So, all of the possible equations are covered in x+1 = 2(x-1) OR x+1 = -[2(x-1)]

Does that help?

Cheers
Brent

### Yes! That makes perfect sense

Yes! That makes perfect sense. Thank you for the speedy response!

### Hello Brent,

Hello Brent,

Need help with these kind of questions that I have no idea how to solve. :(

Q1. If y = |x+5| − |x−5|, then y can take how many integer values?
A. 5
B. 10
C. 11
D. 20
E. 21

Q2. How many solutions are possible for the inequality |x - 1| + |x - 6| < 2?
A. 3
B. 2
C. 1
D. 0
E. 4

Q3. If f(x) = |4x - 1| + |x - 3| + |x + 1|, what is the minimum value of f(x)?
A. 3
B. 4
C. 5
D. 21/4
E. 7

### Question #1: Here's my very

Question #1: Here's my very detailed solution: https://gmatclub.com/forum/if-y-x-5-x-5-then-y-can-take-how-many-integer...

Question #2: Mitch (aka GMATGuruNY) uses a similar strategy to solve the question here: https://www.beatthegmat.com/how-many-solutions-are-possible-for-the-ineq...

Question #3: Mitch (aka GMATGuruNY) has a great solution here: https://www.beatthegmat.com/please-elaborate-basis-number-line-t272128.html

Note: If you need me to elaborate on any parts of Mitch's solutions, just let me know.

Cheers,
Brent

### in questions like

In questions like |x+3| = 2|x-3|, do we have to solve both of the absolute values or only one?
Do we have make all cases?
Like +-(x+3) = +-2(x-3)

### I think you're asking whether

I think you're asking whether we need to examine all 4 possible cases:
i) (x + 3) = 2(x - 3)
ii) -(x + 3) = 2(x - 3)
iii) (x + 3) = -[2(x - 3)]
iv) -(x + 3) = -[2(x - 3)]

The answer to your question is "No, we need only examine 2 cases (cases i and ii)"

The reason for this is that cases i and iv are IDENTICAL, and cases ii and iii are IDENTICAL.
Here's what I mean:

If we take case i and multiply both sides of the equation by -1, we get the EQUIVALENT equation -(x + 3) = -[2(x - 3)], which is IDENTICAL to case iv

Likewise, if we take case ii and multiply both sides of the equation by -1, we get the EQUIVALENT equation (x + 3) = -[2(x - 3)], which is IDENTICAL to case iii

Does that help?

Cheers,
Brent

### If x is an integer, how many

If x is an integer, how many possible values of x exist for
x^2+5|x|+6=0?

A. 4
B. 2
C. 3
D. 1
E. 0
in this i did this
x>=0
then x^2 +5x+6 = 0
x= -2,-3
but x>=0 so -2,-3 rejected
case 2
x<0
x^2-5x+6= 0
x= 2,3
but x<0 so 2,3 rejected
hence no solution
is this approach correct?

### That's a valid solution.

That's a valid solution.
Here's my alternate solution: https://gmatclub.com/forum/if-x-is-an-integer-how-many-possible-values-2...

Cheers,
Brent

### Question;

Question;
If |x| = |y|, is x/y = -1?

(1) x < 0
(2) y > 0

Hi Brent, from the video, you mentioned that the |-3| = 3 which is the same as |-x| = x if x < 0; if this is so, how did you arrive at the solution in Case A

if x < 0, say -2; shouldn't the absolute value be 2?

Below is you solution to Statement 1:

Statement 1: x < 0
There's no information about y, so this statement probably isn't sufficient. let's TEST some values.
There are several values of x and y that satisfy statement 1. Here are two:
Case a: x = -2 and y = 2. Notice that |x| = |y|. In this case x/y = (-2)/2 = -1
Case b: x = -2 and y = -2. Notice that |x| = |y|. In this case x/y = (-2)/(-2) = 1
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Be careful. You wrote: |-3| = 3 which is the same as |-x| = x if x < 0

This is not true.
|-x| = x if x is GREATER THAN 0 (not less than)
For example, if x = 7, we get: |-7| = 7...perfect!!

Let's examine Case a...
Case a: x = -2 and y = 2.
Notice that |x| = |y|, because |-2| = 2 and |2| = 2
In this case x/y = (-2)/2 = -1

NOTE: When we evaluate x/y, there are no absolute values to calculate. We're simply dividing the x-value (-2) by the y-value (2)

Does that help?

Cheers,
Brent

### https://gmatclub.com/forum

https://gmatclub.com/forum/m15-184069.html

### Here's my solution: https:/

Here's my solution: https://gmatclub.com/forum/m15-184069-20.html#p2212604

Cheers,
Brent

### Hi Brent,

Hi Brent,

Cannot understand why statement 2 is not sufficient in this one:

If ab ≠ 0, is |a – b| > |a + b| ?

(1) ab < 0
(2) a > b

Snt 2: I took two values of a and b, both a and b as negative and as positive concluding it to be sufficient.

### Did you try one positive and

Did you try one positive and one negative?

Statement 2: a > b

Case a: a = 5 and b = -1
So, |a – b| = |5 – (-1)| = |6| = 6
And |a + b| = |5 + (-1)| = |4| = 4
Here, the answer to the target question is YES, |a – b| IS greater than |a + b|

Case a: a = 5 and b = 1
So, |a – b| = |5 – 1| = |4| = 4
And |a + b| = |5 + 1| = |6| = 6
Here, the answer to the target question is NO, |a – b| is NOT greater than |a + b|

Cheers,
Brent

### Silly mistake! Nope I did not

Silly mistake! Nope I did not. I have been facing this issue a lot, of not trying all the possible cases.

### Hi Brent, the below is a

Hi Brent, the below is a unique one for me till date.
Q : https://gmatclub.com/forum/what-is-the-value-of-x-100526.html

Statement 1- gives 2 values of x, so the target question has 2 solutions

Statement 2- give 2 values of x, so the target question has 2 solutions.

But when combined, the target question had one common solution with both statements so sufficient.

So to summarize is this also a way to say the data is sufficient, a target question has multiple solutions but if there is a solution which is common from both statements, that's the solution and hence sufficient?

That's correct; if there's exactly one solution in common with both statements, then the combined statements are sufficient.

Cheers,
Brent

Ok Thanks!

### https://www.beatthegmat.com

https://www.beatthegmat.com/if-x-is-an-integer-how-many-possible-values-t299025.html hi Brent! Could you please solve it using the technique showed in your video once please!

### https://www.beatthegmat.com

As you can see from my solution, we can find the answer quickly by applying some number sense. That said, here's the solution using "traditional" methods.

Given: x² + 5|x| + 6 = 0
Start solving for |x|.....
So, 5|x| = -x² - 6
And |x| = (-x² - 6)/5

So, we get two equations:
x = (-x² - 6)/5 and x = -(-x² - 6)/5

Multiply both sides by 5 to get: 5x = -x² - 6
Rearrange to get: x² + 5x + 6 = 0
Factor: (x + 2)(x + 3) = 0
Solve: x = -2 or x = -3

Looks promising!!!
However, we must always test the solutions for EXTRANEOUS ROOTS!

If x = -2, we get: (-2)² + 5|-2| + 6 = 0
Evaluate to get: 4 + 10 + 6 = 0. Doesn't work.

If x = -3, we get: (-3)² + 5|-3| + 6 = 0
Evaluate to get: 9 + 15 + 6 = 0. Doesn't work.

-----------------------------
Now try: x = -(-x² - 6)/5
This is the same as: x = (x² + 6)/5
Multiply both sides by 5 to get: 5x = x² + 6
Rearrange to get: x² - 5x + 6 = 0
Factor: (x - 2)(x - 3) = 0
Solve: x = 2 or x = 3

Looks promising!!!
However, we must always test the solutions for EXTRANEOUS ROOTS!

If x = 2, we get: 2² + 5|2| + 6 = 0
Evaluate to get: 4 + 10 + 6 = 0. Doesn't work.

If x = 3, we get: 3² + 5|3| + 6 = 0
Evaluate to get: 9 + 15 + 6 = 0. Doesn't work.

So, there are NO SOLUTIONS to the original equation.

Cheers,
Brent

### Dear Brent,

Dear Brent,

could you please explain me the solution of:

If k^2 = m^2, which of the following must be true?

(A) k = m
(B) k = −m
(C) k = |m|
(D) k = −|m|
(E) |k| = |m|

I can't wrap my mind around it.

Thanks!

### Hi Brent ,

Hi Brent ,

https://gmatclub.com/forum/m30-186700.html

I can not understand how to solve such questions. Please tell me the process wherein there are multiple absolute values in one equation . Thankyou!

### This is a very tricky

This is a very tricky question. It relies on our understanding that |x-c| represents the DISTANCE between x and c on the number line.

Here's my full solution: https://gmatclub.com/forum/if-y-x-5-x-5-then-y-can-take-how-many-integer...

### I understand that you divided

I understand that you divided the regions and tested for the values in them by plugging them in the equation , what i dont understand is why did you minimize the value of |x+5| and |x-5|?

### The critical points in that

The critical points in that particular question are similar to the critical points we find in quadratic inequalities.
So, by minimizing the values of |x+5| and |x-5|, which happen to be minimized what x+5 and x-5 equal ZERO.

Does that help?

Cheers,
Brent

### Ahh , so even if the equation

Ahh , so even if the equation doesnt equate to 0 , thats how we find the critical points . Thanks , Brent !

### Yes, when it comes to

Yes, when it comes to absolute values (in this case)

### Hi Brent,

Hi Brent,

Unfortunately, absolute value is not my strongest point in math. I remember I did not fully understand this in school, so here is my second chance I guess. Here is my question.
In your video you explain that |y| = y, when y>0, and |y|= -y when y<0.
In the question above in your explanation, you have |-4|=4 and I am puzzled here if y=-4, and we know that -4 is less than 0, then why |-4| does not equal to -4, but is equal to 4 instead. What am I missing here?

Thanks!

You're correct to say that, |y|= -y when y < 0 (i.e., y is negative)

IMPORTANT: Keep in mind, that if y is negative, then -y will be positive.
For example, if y = -1, them -y = -(-1) = 1

Here are some examples, when y (the part inside the absolute value symbols) is negative:
|-3| = -(-3) = 3
|-11| = -(-11) = 11
|-8| = -(-8) = 8

Likewise, when y = -4 (y is negative), we get: |-4| = -(-4) = 4

Does that help?

Cheers,
Brent

### Hi Brent,

Hi Brent,

question https://gmatclub.com/forum/what-is-the-sum-of-all-solutions-to-the-equation-x-4x-228473.html

From the question above I quickly realized that |x^2-4x+4| is equal to |x-2|^2 and x^2+10x-24 = (x+12)(x-2), hence we will have |x-2|^2=(x+12)(x-2), we can quickly realize that one solution to this equation is x=2, and then we divide both sides by (x-2) we can solve can solve x-2 = -x-12, solve equation and get a root x=-5, obviously we have to test in the original equation. Is it a valid approach? Is there such an approach or the best way is to always go the standard way outlined in your solution?

Thanks!

Your strategy works for that particular question, but it's hard to see why, in the second part, you chose to solve x-2 = -x-12. (I know why you did that. So, as long as you understand, then the approach will always work).

That said, it only works when you can factor the same binomial from each side.
It won't work for something like: |(x+1)(x+2)| = (x-3)(x+5)

Cheers,
Brent

### At 00:22, it's mentioned that

At 00:22, it's mentioned that the absolute value of |x| is negative x, if x is less than zero. I thought absolute values are always positive? Thanks.

### You're correct to say that

You're correct to say that absolute values are always greater than or equal to zero.
However, -x is not necessarily negative.
Notice that, when x is a negative number, then -x is a positive number.
For example, if x equals -5, then -x = -(-5) = 5

At 0:22 point in the video, I make the following observations:
If x ≥ 0, then |x| = x
For example, if x = 2, then |2| = 2

If x < 0, then |x| = -x
For example, if x = -6, then |-6| = -(-6) (which equals 6)

Does that help?

Cheers,
Brent

### Yes. I just got confused.

Yes. I just got confused. Thanks.

### If y + |y| = 0 which of the

If y + |y| = 0 which of the following must be true?

A. y > 0
B. y ≥ 0
C. y < 0
D. y ≤ 0
E. y = 0
———————————————
I strongly believe the question has multiple correct answers option C, D, E are all correct

Test: Y + |Y| = 0

C. Y = -ive —> Eg: -2
-2 + |-2| = 0
-2 + 2 = 0
0 = 0 ——> True
———————————
E. Y + |Y| = 0
0 + |0| = 0
0 + 0 = 0
0 = 0 —-> True

—————————————
D. Y less than equal to 0
Both Y = 0 & Y = less than 0[ => i.e. -ive value]
Has been proven ——> True

### If the question asked "Which

If the question asked "Which of the following COULD be true?", then you're correct to say that C, D and E would be correct.

However, the word MUST significantly changes things.

Take answer choice C, for example.
You've already shown that y COULD be zero, in which case, we can't say that y < 0 MUST be true.

The same applies to answer choice E.
We already know that y = -1 is a possible solution to the equation y + |y| = 0
So, we can't say that y = 0 MUST be true

Cheers,
Brent

### https://www.beatthegmat.com

https://www.beatthegmat.com/value-of-y-t279564.html

I need help with the statement 1 of the question linked above.
1.) Why x’s value were assumed as 0 and 2?
2.) How to solve statement 1 step by step( only when not assumed)?

1) x = 0 and x = 2 are just two possible solutions. As I mentioned in my solution (link below), there are infinitely many solutions to the equation given in statement 1.

2) 3|x² – 4| = y – 2
Noticed that this equation has two variables.
This means that each solution consists of an x-value and a y-value.
If we add 2 to both sides of this equation we get 3|x² – 4| + 2 = y

In other words y = 3|x² – 4|+ 2
So, for example, if x = 1, then y = 3|1² – 4| + 2 = 11
If x = 2, then y = 3|2² – 4| + 2 = 2
If x = 3, then y = 3|3² – 4| + 2 = 17
If x = 4, then y = 3|4² – 4| + 2 = 38
If x = 0, then y = 3|0² – 4| + 2 = 14
etc...

Here's my full solution: https://gmatclub.com/forum/what-is-the-value-of-y-127976-40.html#p2452306

Cheers,
Brent

Thanks!