The Algebra Teacher\'s Guide to Reteaching Essential Concepts and Skills

Teaching Notes 3.16: Solving Absolute Value Inequalities

An absolute value inequality may be equivalent to a conjunction or a disjunction, depending on

how the absolute value compares to a positive number. Many students do not realize this and

consequently have trouble solving these inequalities.

1. Explain that the first steps to solving an absolute value inequality are to isolate the absolute
   value expression and then write a correct inequality.


2. Explain that an absolute value inequality may be equivalent to a conjunction or a disjunc-
   tion. If the expression within the absolute value symbols is< or≤a positive number, it is
   equivalent to a conjunction. If the expression within the absolute value symbols is>or≥a
   positive number, it is equivalent to a disjunction.


3. To show why some inequalities are equivalent to a conjunction, present this example:|x|<
   7. Ask your students to substitute 3,−1, 8, 6,−6, and−7forxto see if the inequality is true.
   They should find that 3,−1, 6, and−6 will make the statement true and conclude thatx< 7
   andx>−7, which can be rewritten as− 7 <x<7.


4. To show why some inequalities are equivalent to a disjunction, present this example:|x|>7.
   Ask your students to substitute 8, 6,−8, 3, 15, and−5forxto see if the inequality is true.
   They should find that 8,−8, and 15 will make the statement true and conclude thatx>7or
   x<−7, which is a disjunction.


5. Review the information and examples on theworksheet with your students. Make sure that
   your students understand the steps for rewriting and solving the equivalent inequalities.

EXTRA HELP:

Absolute value inequalities will always be>,≥,<,or≤a positive number.

ANSWER KEY:

(1)D;y>1ory<− 13 (2)C;x<1andx>−7;− 7 <x< 1

(3)C;y≤−4andy≥−8;− 8 ≤y≤− 4 (4)D;y>2ory<− 2

(5)D;y≥11 ory≤− 12 (6)C;x<6andx>−1;− 1 <x< 6

(7)D;x>48 orx<− 48 (8)C;y≥−5andy≤15;− 5 ≤y≤ 15

------------------------------------------------------------------------------------------

(Challenge)Meg made her first mistake when she wrote 2x<−4 instead of 2x>−4.The solution

should bex<2andx>−2, which can be rewritten as− 2 <x<2.

------------------------------------------------------------------------------------------

118 THE ALGEBRA TEACHER’S GUIDE