Absolute Value Inequalities

• Solve a compound inequality such as −3 < 2x + 1 ≤ 7 by doing the same operation to all three parts
• Solve |expression| < c as an AND statement: −c < expression < c (one connected interval)
• Solve |expression| > c as an OR statement: expression < −c or expression > c (two separate rays)
• Match an absolute value inequality to its number-line graph, with open or closed endpoints
• Handle the special cases: |expression| < a negative number has no solution, and |expression| > a negative number is true for all real numbers

A compound inequality puts a variable expression between two bounds. Solve it by doing the same operation to all three parts: −3 < 2x + 1 ≤ 7 becomes −4 < 2x ≤ 6 after subtracting 1, then −2 < x ≤ 3 after dividing by 2. As always, dividing or multiplying by a negative flips every inequality sign. Absolute value measures distance from zero, which gives two patterns for c > 0. Less than means 'within c of zero,' an and statement: |x| < c means −c < x < c, one connected interval. Example: |2x − 1| ≤ 5 means −5 ≤ 2x − 1 ≤ 5, so −4 ≤ 2x ≤ 6 and −2 ≤ x ≤ 3. Greater than means 'more than c from zero,' an OR statement: |x| > c means x < −c or x > c, two rays pointing away from each other. Example: |x + 2| > 3 means x + 2 < −3 or x + 2 > 3, so x < −5 or x > 1. On a number line, < and > use open circles and ≤ and ≥ use closed circles; a 'less than' absolute value inequality shades between two points and a 'greater than' one shades outside them. The ACT also runs this backwards: 'all values within 3 units of 5' is |x − 5| ≤ 3, where 5 is the midpoint of the interval and 3 is half its width. So the interval 2 ≤ x ≤ 8 is |x − 5| ≤ 3. Special cases: an absolute value is never negative, so |x − 4| < −2 has no solution, while |x − 4| > −2 is true for every real number. Isolate the absolute value before splitting: 2|x| − 3 < 7 becomes |x| < 5 first.

• Less than is and (between). Greater than is OR (outside). Say which one it is before you write anything.
• Isolate the absolute value bars first, exactly as you would isolate a variable.
• To write an interval as an absolute value inequality, find the midpoint and the half-width: |x − midpoint| ≤ half-width.
• Test one number from your answer in the original inequality. It takes five seconds and catches a flipped sign.