Absolute Value Equations Topic Index | Algebra2/Trig Index | Regents Exam Prep Center

To solve an absolute value equation, isolate the absolute value on one side of the equal sign, and establish two cases:

 Case 1:               Set the expression inside the              absolute value symbol equal to              the other given expression. Case 2:               Set the expression inside the              absolute value symbol equal to              the negation of the other given              expression . . .  and always CHECK your answers. The two cases create "derived" equations.  These derived equations may not always be true equivalents to the original equation.  Consequently, the roots of the derived equations MUST BE CHECKED in the original equation so that you do not list extraneous roots as answers.

Your graphing calculator can be used to solve absolute value equations and/or double check your answers.

 How to use your TI-83+/84+ graphing calculator  with absolute value equations. Click calculator.

Example 1:
(Two cases with two solutions)

 Case 1: Case 2: Answer:  x = 16,  x = 4 Check: Check: The solutions are 16 or 4.  On a number line, these value are each 6 units away from 10.

 Remember:  Absolute value is always positive (or zero).  An equation such as  is never true.  It has NO solution.  The answer is the empty set .

Example 2:
(No solution)

As soon as you isolate the absolute value expression, you observe:

There is no need to work out the two cases in this problem.  Absolute value is NEVER equal to a negative value.  This equation is never true.  The answer is the empty set .

Example 3:             (Two cases with one solution)

 Case 1: Case 2: Answer:  x = -1 Check: Not an answer! Check: The check shows that x = -3 is NOT a solution to this absolute value equation.  There is only ONE answer, x = -1. Always check!!!!

Example 4:   A machine fills Quaker Oatmeal containers with 32 ounces of oatmeal.  After the containers are filled, another machine weighs them.  If the container's weight differs from the desired 32 ounce weight by more than 0.5 ounces, the container is rejected.  Write an equation that can be used to find the heaviest and lightest acceptable weights for the Quaker Oatmeal container.  Solve the equation.

Solution:  Let x = the weight of the container.

 Case 1: Case 2:

Answer:  x = 31.5 ounces (lightest)
x = 32.5 ounces (heaviest)

When setting up a word problem involving absolute value, remember that absolute value can represent "distance" from a given point.

The difference between the answer (x) and the desired point (32) is placed under the absolute value symbol.  This absolute value is then set equal to the desired "distance" (0.5).

 Topic Index | Algebra2/Trig Index | Regents Exam Prep Center Created by Donna Roberts