Lesson: Absolute Value Equations

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?
thanks
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
NO GOOD.
So, y cannot equal 2.
ELIMINATE A and B

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

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

By the process of elimination, the correct answer is D

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