Solving Absolute Value Equations and Inequalities Section 2.8 Solving Absolute Value Equations and Inequalities

Homework Pg #20-27, 32-35, 59, 60

A compound statement is made up of more than one equation or inequality. A disjunction is a compound statement that uses the word or. Disjunction: x ≤ –3 OR x > 2 Set builder notation: {x|x ≤ –3 U x > 2} A disjunction is true if and only if at least one of its parts is true.

A conjunction is a compound statement that uses the word and. Conjunction: x ≥ –3 AND x < 2 Set builder notation: {x|x ≥ –3 x < 2}. A conjunction is true if and only if all of its parts are true. Conjunctions can be written as a single statement as shown. x ≥ –3 and x< 2 –3 ≤ x < 2 U

Solving Compound Inequalities Solve the compound inequality. Then graph the solution set. 6y < –24 OR y +5 ≥ 3 Solve both inequalities for y. 6y < –24 y + 5 ≥3 or y < –4 y ≥ –2 The solution set is all points that satisfy {y|y < –4 or y ≥ –2}. –6 –5 –4 –3 –2 –1 0 1 2 3 (–∞, –4) U [–2, ∞)

Solving Compound Inequalities Solve the compound inequality. Then graph the solution set. Solve both inequalities for c. and 2c + 1 < 1 c ≥ –4 c < 0 The solution set is the set of points that satisfy both c ≥ –4 and c < 0. –6 –5 –4 –3 –2 –1 0 1 2 3 [–4, 0)

Recall that the absolute value of a number x, written |x|, is the distance from x to zero on the number line. Because absolute value represents distance without regard to direction, the absolute value of any real number is nonnegative.

Absolute-value equations and inequalities can be represented by compound statements. Consider the equation |x| = 3. The solutions of |x| = 3 are the two points that are 3 units from zero. The solution is a disjunction: x = –3 or x = 3.

The solutions of |x| < 3 are the points that are less than 3 units from zero. The solution is a conjunction: –3 < x < 3.

The solutions of |x| > 3 are the points that are more than 3 units from zero. The solution is a disjunction: x < –3 or x > 3.

Note: The symbol ≤ can replace <, and the rules still apply Note: The symbol ≤ can replace <, and the rules still apply. The symbol ≥ can replace >, and the rules still apply.

Solving Absolute-Value Equations Solve the equation. This can be read as “the distance from k to –3 is 10.” |–3 + k| = 10 Rewrite the absolute value as a disjunction. –3 + k = 10 or –3 + k = –10 Add 3 to both sides of each equation. k = 13 or k = –7

Solving Absolute-Value Equations Solve the equation. Isolate the absolute-value expression. Rewrite the absolute value as a disjunction. Multiply both sides of each equation by 4. x = 16 or x = –16

Solving Absolute-Value Inequalities with Disjunctions Solve the inequality. Then graph the solution. |–4q + 2| ≥ 10 Rewrite the absolute value as a disjunction. –4q + 2 ≥ 10 or –4q + 2 ≤ –10 Subtract 2 from both sides of each inequality. –4q ≥ 8 or –4q ≤ –12 Divide both sides of each inequality by –4 and reverse the inequality symbols. q ≤ –2 or q ≥ 3

Solving Absolute-Value Inequalities with Disjunctions Solve the inequality. Then graph the solution. |0.5r| – 3 ≥ –3 Isolate the absolute value as a disjunction. |0.5r| ≥ 0 Rewrite the absolute value as a disjunction. 0.5r ≥ 0 or 0.5r ≤ 0 Divide both sides of each inequality by 0.5. r ≤ 0 or r ≥ 0 The solution is all real numbers, R. –3 –2 –1 0 1 2 3 4 5 6 (–∞, ∞)

Solving Absolute-Value Inequalities with Conjunctions Solve the compound inequality. Then graph the solution set. |2x +7| ≤ 3 Multiply both sides by 3. Rewrite the absolute value as a conjunction. 2x + 7 ≤ 3 and 2x + 7 ≥ –3 Subtract 7 from both sides of each inequality. 2x ≤ –4 and 2x ≥ –10 Divide both sides of each inequality by 2. x ≤ –2 and x ≥ –5

Solving Absolute-Value Inequalities with Conjunctions Solve the compound inequality. Then graph the solution set. Multiply both sides by –2, and reverse the inequality symbol. |p – 2| ≤ –6 Rewrite the absolute value as a conjunction. |p – 2| ≤ –6 and p – 2 ≥ 6 Add 2 to both sides of each inequality. p ≤ –4 and p ≥ 8 Because no real number satisfies both p ≤ –4 and p ≥ 8, there is no solution. The solution set is ø.