| Absolute Values |

by on September 14th, 2009

Absolute Values (AVs) are confusing. They seem so simple, but the questions end up being so hard. Why is that? Here we will address a few rules to keep in mind when studying for the GMAT.

1. Absolute Value equations are two equations disguised as one

You can split up any equation involving absolutes into two, and solve for each solution. One will look identical to the given, and the other is found by multiplying the inside by -1. Remember to multiply the entire expression by -1.

| (x + 5)/3 | = 11 turns into:

  • (x + 5)/3 = 11, and
  • (x+5)/3 = -11

x + 5 = 33

x + 5 = -33

x = 28 x = -38

Note that plugging either x = 28 or x = -38 into the original equation will check out. Also note that solutions for variables within absolute value questions can be negative. What is spit out of the AV cannot be negative, but what goes in can be anything.

2. Think of Absolute Values as distances from zero

If an AV = 15, that means whatever is inside the AV is exactly 15 above or below zero on the number line.

| x + 5 | = 15

_____-20_____-15_____-10__________0__________+10_____+15_____+20_____

x = -20 and x = 10. Note that this is a SHIFT of -5 from the constant on the right side of the equation (15). If the variable inside the AV is being added, shift to the left to find the solution. If the variable inside the AV is being subtracted, shift to the right to find the solution.

3. Thank about your range of options

If posed with an inequality, think about the range of options that would satisfy it. It may help to pick a number and plug in to make sure you haven’t mixed it up.

| x + 5 | < 15

____________-15_______________0_______________+15___________

The only difference between | x + 5 | < 15 and | x + 5 | = 15 is that you are looking for a range of x, not what it equals. However, you can solve for the extremes (-20, +10 from previous example) and then look at the sign. Since | x + 5 | is LESS than 15, that means | x + 5 | is within 15 of 0. So your range is -20 < x < 10.

Hypothetically, if we flipped the sign to | x + 5 | > 15, your satisfying values would be x < -20 and x > 10.

Examples

If | x | + | y | = | x + y |, which of the following statements must be true?

(A) xy > 0

(B) xy < 0

(C) x + y > 0

(D) x + y < 0

(E) x – y > 0

Again, keep in mind that everything coming out of an AV is positive, but not necessarily everything that goes in. If both | x | + | y | and | x + y | must be positive, we must look at how the input values affect the output. The absolute values add on top of each other on the | x | + | y | side in positive space. This means that the x and y within | x + y | must add on top of each other in the same direction. To do this, x and y must be the same sign, but says nothing about their relative sizes. If x and y are the same sign. xy > 0. Choice (A).

What is x? (Data Sufficiency)

(1) | x | < 2

(2) | x | = 3x – 2

As discussed above, AV equations give 2 solutions and AV inequalities give 2 endpoints of ranges. In this sense, we know that neither Statement (1) nor Statement (2) is sufficient, right? Not so fast.

  • Statement (1) gives us a range, and tells us that -2 < x < 2.
  • Per strategy above, we can split Statement (2) into the following two equations:

x = 3x -2,  and

-x = 3x – 2

x =1

x = 1/2

It seems that even with (1) and (2) TOGETHER,  we still would not know whether x = 1 or x = 1/2. However, these questions emphasize the need of plugging back into the original equation to check your answer.

If we were to plug in x = 1/2 into | x | = 3x – 2, we would get 1/2 = -1/2, which is NOT true. Since this solution cannot exist, only x = 1 is our solution. Answer Choice (B).

Be extra careful when making +/- assumptions when there is a variable on both sides of the equation. Graphically, outputs of AV expressions cannot be below the x-axis. In this case, the line y = 3x – 2 hits the x-axis at x= 2/3. Any x value below 2/3 cannot be a solution to the given equation, since that would imply that | x | was negative.

Any more tips? Just post in the comment field below to discuss? Good luck!

3 comments

  • Thanks Jake !
    Thanks for the tip..

    ~ Vivek

  • Hey Jake,
    Great article, luved it.
    There is one thing which i still need to know. In regards to the DS question for absolute values, do we have to always plug in the value and cross verify? If not, then how do we come to know if it is required or not.

    Thanks
    Ishaan

  • Couldn't the last example have been solved in the following way:

    |x|=3x-2
    x = 3x - 2
    x = -3x + 2
    3x - 2 =-3x + 2
    6x=4
    x=2/3

    I can actually see that this result cannot fit as a right solution in the original equation.

    Could you point out what is wrong with this logic, please?

    And why when doing the same thing for |x+5|=15 (=> x+5=15, x+5=-15), the result is finally the correct solution for the equation?

    Thank you in advance!

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