# 1.4 – Solving Absolute Value Equations and Inequalities

Objectives

• Identify problems that require the use of absolute value.
• Transform absolute value problems into compound inequalities or equations.

Key Terms

• Absolute Value – a number’s distance from zero on a number line. ex. |3| = 3 and |-3| = 3 because both 3 and -3 are a total of 3 real spaces away from zero.
• Intersection – the set of numbers that appear in the solution sets of two inequalities.
• Union – the set of all solutions that are found in an or statement.

Notes

• Solve: |x| = 2

Answer: x represents a number 2 spaces from zero. It can be negative or positive; so, x = 2 OR x = -2. Both are correct answers.

• Steps to Solve an Absolute Value Equation

Step 1: If necessary, get the absolute value expression by itself on one side of the equation.

Step 2: Rewrite the absolute value equation as two separate equations: one positive and the other negative.

Step 3: Solve each equation separately.

Step 4: After solving, substitute your answers back into the original equation to verify that your solutions are valid.

• ex. |2x + 1| = 5
• Step 1: skip
• Step 2:
• a. 2x + 1 = 5
• b. -(2x + 1) = 5
* VERY IMPORTANT: the one that is negative NEEDS to have the negative sign outside of the parenthesis. You WILL distribute the negative sign to solve it in the next step.
• Step 3:
• a. 2x + 1 = 5 (subtract 1 on both sides)
2x = 4 (then divide by 2 on both sides)
x = 2
• b. -(2x + 1) = 5 (distribute the negative sign into the parenthesis)
-2x – 1 = 5 (add one to both sides)
-2x = 6 (then divide by -2 on both sides)
x = -3
• Step 4: Substitute the answers for x into the original equation and check for validity
• a. |2(2) + 1| = 5
|4 + 1| = 5
|5| = 5 (true)
• b. |2(-3) + 1| = 5
|-6 + 1| = 5
|-5| = 5 (true)

• No Solution
• a. Absolute value is ALWAYS positive, so there is no solution for |x| = -a (“a” is any real number) because absolute values can NEVER be negative.
• ex. 2•|3x – 9| = -6
• Divide both sides by 2 and see if the right side is still a negative: |3x – 9| = -3
• Since the right side is -3 (a negative integer), the absolute value has NO solutions because it can’t equal a negative number.
• b. When you substitute answers back in to check for validity, one may not work out.
• ex. |x – 3| = x + 2
• a. x – 3 = x + 2 (subtract x on both sides)
-3 = 2 (false, impossible: no solution)
• b. -(x – 3) = x + 2 (distribute the negative sign)
-x + 3 = x + 2 (add x to both sides)
3 = 2x + 2 (subtract 2 on both sides)
1 = 2x (divide by 2 on both sides)
1/2 = x (true, one solution)
• Absolute Value Inequalities: Greater Than Signs ( > ): OR
• The solution to the inequality is EITHER / OR, not both (so do NOT use the word: AND).
• Step 1: |x| > 3
• Remember to set this up as a positive and negative (using parenthesis):
• a. x > 3
• b. -(x) > 3 … so, -x > 3 (divide by -1 on both sides and FLIP the sign)
• Step 2:  Rewrite the answer where x is positive:  x < -3  OR  x > 3
• Step 3: Graph

• Absolute Value Inequalities: Less Than Signs ( < ): AND
• The solution to the inequality is between the positive and negative values: AND (you can write “or,” but AND is more appropriate).
• Step 1: |x| > 3
• Remember to set this up as a positive and negative (using parenthesis):
• a. x < 3
• b. -(x) < 3 … so, -x < 3 (divide by -1 on both sides and FLIP the sign)
• Step 2:  Rewrite the answer where x is positive:  x < 3  AND  x > -3, but the best way to write this is: -3 < x < 3
• You will sometimes still see this written with OR: x < 3 or x > -3, but if you graph it, you will see why it is AND.
• Step 3: Graph

• Cheat Sheet: If “a” is any real number:
• |x| < -a is NEVER true (no solution) b/c absolute value is always positive
• |x| > -a is ALWAYS true (all real numbers) b/c absolute value is always positive and positive is greater than negative.
• |x| > +a is SOMETIMES true (some solutions will work, but not others)
• |x| < +a is ALWAYS true (solutions will be between -a and +a values)

• Real World Example
• The ideal length of a particular metal rod is 20.5 cm.
• The measured length may vary from the ideal length by at most 0.045 cm.
• What is the range of acceptable lengths for the rod?
• Setup:  | x – 20.5 | ≤ 0.045 (use a negative sign in the setup b/c you are looking for the difference!
• Create both inequalities (positive and negative):
• a. (x – 20.5) ≤ 0.045
• b. – (x – 20.5) ≤ 0.045 which is simplified to:  -x + 20.5 ≤ 0.045
• Solve both inequalities with an AND result
• So, the range is between the two values: 20.455  ≤  x  ≤  20.545