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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.
