Holt Algebra 2 2-8 Solving Absolute-Value Equations and Inequalities Section 2.8 Solving Absolute Value Equations and Inequalities.

Holt Algebra 2

2-8Solving Absolute-Value Equations and Inequalities

Section 2.8

Solving Absolute Value Equations and Inequalities

Holt Algebra 2

2-8Solving Absolute-Value Equations and Inequalities

Homework• Pg #20-27, 32-35, 59, 60

Holt Algebra 2

2-8Solving Absolute-Value Equations and Inequalities

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.

Holt Algebra 2

2-8Solving Absolute-Value Equations and Inequalities

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

Holt Algebra 2

2-8Solving Absolute-Value Equations and Inequalities

Solving Compound Inequalities

Solve the compound inequality. Then graph the solution set.

Solve both inequalities for y.

The solution set is all points that satisfy {y|y < –4 or y ≥ –2}.

6y < –24 OR y +5 ≥ 3

6y < –24 y + 5 ≥3

y < –4 y ≥ –2or

–6 –5 –4 –3 –2 –1 0 1 2 3 (–∞, –4) U [–2, ∞)

Holt Algebra 2

2-8Solving Absolute-Value Equations and Inequalities

Solve both inequalities for c.

The solution set is the set of points that satisfy both c ≥ –4 and c < 0.

c ≥ –4 c < 0

–6 –5 –4 –3 –2 –1 0 1 2 3[–4, 0)

Solving Compound Inequalities

Solve the compound inequality. Then graph the solution set.

and 2c + 1 < 1

Holt Algebra 2

2-8Solving Absolute-Value Equations and Inequalities

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.

Holt Algebra 2

2-8Solving Absolute-Value Equations and Inequalities

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.

Holt Algebra 2

2-8Solving Absolute-Value Equations and Inequalities

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

Holt Algebra 2

2-8Solving Absolute-Value Equations and Inequalities

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.

Holt Algebra 2

2-8Solving Absolute-Value Equations and Inequalities

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

Holt Algebra 2

2-8Solving Absolute-Value Equations and Inequalities

Solve the equation.

Solving Absolute-Value Equations

Rewrite the absolute value as a disjunction.

This can be read as “the distance from k to –3 is 10.”

Add 3 to both sides of each equation.

|–3 + k| = 10

–3 + k = 10 or –3 + k = –10

k = 13 or k = –7

Holt Algebra 2

2-8Solving Absolute-Value Equations and Inequalities

Solve the equation.

Solving Absolute-Value Equations

x = 16 or x = –16

Isolate the absolute-value expression.

Rewrite the absolute value as a disjunction.

Multiply both sides of each equation by 4.

Holt Algebra 2

2-8Solving Absolute-Value Equations and Inequalities

Solving Absolute-Value Inequalities with Disjunctions

Solve the inequality. Then graph the solution.

Rewrite the absolute value as a disjunction.

|–4q + 2| ≥ 10

–4q + 2 ≥ 10 or –4q + 2 ≤ –10

–4q ≥ 8 or –4q ≤ –12

Divide both sides of each inequality by –4 and reverse the inequality symbols.

Subtract 2 from both sides of each inequality.

q ≤ –2 or q ≥ 3

Holt Algebra 2

2-8Solving Absolute-Value Equations and Inequalities

Solve the inequality. Then graph the solution.

|0.5r| – 3 ≥ –3

0.5r ≥ 0 or 0.5r ≤ 0

Divide both sides of each inequality by 0.5.

Isolate the absolute value as a disjunction.

r ≤ 0 or r ≥ 0

|0.5r| ≥ 0

Rewrite the absolute value as a disjunction.

Solving Absolute-Value Inequalities with Disjunctions

–3 –2 –1 0 1 2 3 4 5 6

(–∞, ∞)

The solution is all real numbers, R.

Holt Algebra 2

2-8Solving Absolute-Value Equations and Inequalities

Solve the compound inequality. Then graph the solution set.

Solving Absolute-Value Inequalities with Conjunctions

|2x +7| ≤ 3 Multiply both sides by 3.

Subtract 7 from both sides of each inequality.

Divide both sides of each inequality by 2.

Rewrite the absolute value as a conjunction.

2x + 7 ≤ 3 and 2x + 7 ≥ –3

2x ≤ –4 and 2x ≥ –10

x ≤ –2 and x ≥ –5

Holt Algebra 2

2-8Solving Absolute-Value Equations and Inequalities

Solve the compound inequality. Then graph the solution set.

Solving Absolute-Value Inequalities with Conjunctions

|p – 2| ≤ –6Multiply both sides by –2, and reverse the inequality symbol.

Add 2 to both sides of each inequality.

Rewrite the absolute value as a conjunction.

|p – 2| ≤ –6 and p – 2 ≥ 6

p ≤ –4 and p ≥ 8

Because no real number satisfies both p ≤ –4 andp ≥ 8, there is no solution. The solution set is ø.