The GRE adds another level of difficulty to equations and inequalities in the form of absolute value.
The “absolute value” of a number describes how far that number is away from 0. It is the distance between that number and 0 on a number line. The symbol for absolute value is |number|. For instance, the absolute value of −5 is written as |−5|.
Example 1: The absolute value of 5 is 5:
Example 2: The absolute value of −5 is also 5:
When you face an expression like |4 − 7|, treat the absolute value symbol like parentheses. Solve the arithmetic problem inside first, and then find the absolute value of the answer. In this case, 4 − 7 = −3, and −3 is three units from zero, so |4 − 7| = |−3| = 3.
Mark the following expressions as True or False.
|3| = 3
|−3| = −3
|3| = −3
|−3| = 3
|3 − 6| = 3
|6 − 3|= −3
On the GRE, some absolute value equations place a variable inside the absolute value signs:
Example: |y| = 3
What’s the trap here? The trap is that there are two numbers, 3 and −3, that are 3 units away from 0. That means both of these numbers could be possible values for y. So how do you figure that out? Here, you can’t. All you can say is that y could be either the positive value or the negative value; y is either 3 or −3.
When there is a variable inside an absolute value, you should look for the variable to have two possible values. Although you will not always be able to determine which of the two is the correct value, it is important to be able to find both values. Following is a step-by-step process for finding all solutions to an equation that contains a variable inside an absolute value:
|y| = 3 |
Step 1: Isolate the absolute value expression on one side of the equation. In this case, the absolute value expression is already isolated. |
+(y) = 3 or − (y) = 3 |
Step 2: Take what’s inside the absolute value sign and set up two equations. The first sets the positive value equal to the other side of the equation, and the second sets the negative value equal to the other side. |
y = 3 or −y = 3 |
Step 3: Solve both equations. |
y = 3 or y = −3 |
Note: There are two possible values for y. |
Sometimes people take a shortcut and go right to “y equals plus or minus 3.” This shortcut works as long as the absolute value expression is by itself on one side of the equation.
Here’s a slightly more difficult problem, using the same technique:
Example: 6 × |2x + 4| = 30
To solve this, you can use the same approach:
6 × |2x + 4| = 30 |
Step 1: Isolate the absolute value expression on one side of the equation or inequality. |
(2x + 4) = 5 or −(2x + 4) = 5 2x + 4 = 5 or −2x − 4 = 5 |
Step 2: Set up two equations—the positive and the negative values are set equal to the other side. |
2x = 1 or −2x = 9 | Step 3: Solve both equations/inequalities. |
x =
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Note: There are two possible values for x. |
Solve the following equations with absolute values in them.
|a| = 6
|x + 2| = 5
|3y −4| = 17
4|x + | = 18