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SAT Tips from Veritas Prep

SAT Tip: Calculating Absolute Values


SAT Tip: Calculating Absolute Values

Photograph by Ralf Wendrich/Getty Images

This tip on improving your SAT score was provided by Vivian Kerr at Veritas Prep.

The simplest way to think of absolute value is as the distance from a number to zero on a number line. Just like shapes such as squares and triangles can’t have side values that are negative (because that wouldn’t make sense in the real world), absolute values will always be positive. We can’t have a “negative” distance. That’s why |-3| = 3, and why also |3| = 3. Both -3 and 3 are (in the real world) three spaces away from zero on a number line.  Absolute value equations will almost always yield two possible solutions.

To solve an equation with an absolute value, first simplify the equation so that one side is surrounded by the absolute value bars, then split it into two equations: one without the absolute value and the other side of the equation left as written, and the other without the absolute value and the other side of the equation with the opposite sign.

In this example, notice that we didn’t split the equation until we had divided both sides by 2 to isolate the absolute value symbol. To double-check our work, let’s plug in:

2|4(5) – 7| = 26

2 |20 – 7| = 26

2|13| = 26

26 = 26

And the other one:

2|4(-3/2) – 7| = 26

2|(-12/2) – 7| = 26

2|-6 – 7| = 26

2|-13| = 26

2 x 13 = 26

26 = 26

Correct!

Let’s try an SAT practice problem together from the Official SAT Guide:

If 2|x+3| = 4 and |y+1|/3 = 2, then |x + y| could equal each of the following EXCEPT:

A. 0

B. 4

C. 8

D. 10

E. 12

From our practice, we know that each of these two absolute value equations will have 2 answers, so there will be 4 possible values for “x + y.” Because the question is using the word “except” we know that 4 of the 5 choices will be a possible “x+y,” and one will not, so let’s find the possible values of x and y:

2|x + 3| = 4

|x + 3| = 2

Now we split into: x + 3 = 2 and x + 3 = -2. The possible values for x are -1 and -5.

|y + 1|/3 = 2

|y + 1| = 6

Now we split into: y + 1 = 6 and y + 1 = -6. The possible values for y are 5 and -7.

Therefore, the possible values of |x + y| are:

|-1 + 5| = |4| = 4

|-1 + -7| = |-8| = 8

|-5 + 5| = |0| = 0

|-5 – 7| = |-12| = 12

The only answer choice that is not one of our four options is (D), so that is the correct answer.

Remember, if you have to do a calculation inside the absolute value symbols, do the calculation as if the absolute value doesn’t exist. Then, if you happen to end up with a negative value, make it positive.

For example, if we have |2 – 4| = x, start by simplifying “2-4” and writing it as -2, then we know that x = 2. If we tried to make -4 positive first, before calculating, it would look like |2 + 4| = |6| = 6, which would give us the wrong answer. Remember: calculation first, absolute value second.

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