Lesson: Absolute Value Equations

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Comment on Absolute Value Equations

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?
gmat-admin's picture

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.

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
Can we chose y as + and - ? I chose y as 1 and | y | as -1 so I add 1 + (-1) = 0 . I don't understand why the answer is d .
gmat-admin's picture

"I chose y as 1 and |y| as -1"

If y = 1, then |y| = |1| = 1 (not -1).

NOTE: The absolute value of ANY value is always greater than or equal to 0. In other words, the absolute value of a number CANNOT be negative.

Let's solve the question by testing values.

Try y = 2. Plug into equation to get: 2 + |2| = 0
Evaluate: 4 = 0
So, y cannot equal 2.

Try y = 0. Plug into equation to get: 0 + |0| = 0
Evaluate: 0 = 0
So, y can equal 0.

Try y = -1. Plug into equation to get: -1 + |-1| = 0
Evaluate: 0 = 0

By the process of elimination, the correct answer is D

Hi Brent,
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?

gmat-admin's picture

Question link: https://gmatclub.com/forum/what-is-the-sum-of-all-the-real-values-of-x-f...

Hi Aladdin,

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.


Hi Brent,
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.
gmat-admin's picture

Question link: https://gmatclub.com/forum/if-x-4-what-is-the-range-of-the-solutions-of-...

Hi Aladdin,

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?


Thanks Brent.

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
gmat-admin's picture

Question link: https://gmatclub.com/forum/if-x-and-y-are-positive-integers-what-is-the-...

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?


Question link: https://gmatclub.com/forum/if-x-1-2-x-1-what-is-the-sum-of-the-roots-244841.html

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?

gmat-admin's picture

Question link: https://gmatclub.com/forum/if-x-1-2-x-1-what-is-the-sum-of-the-roots-244...

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?


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

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
gmat-admin's picture

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.


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)
gmat-admin's picture

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?


If x is an integer, how many possible values of x exist for

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

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


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