Text of Holt Algebra 2 2-8 Solving Absolute-Value Equations and Inequalities Solve compound inequalities....
Slide 1Solve compound inequalities.
Objectives
Solving 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
A disjunction is true if and only if at least one of its parts is
true.
Holt Algebra 2
Solving Absolute-Value Equations and Inequalities
A conjunction is a compound statement that uses the word and.
Conjunction: x ≥ –3 AND 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
Holt Algebra 2
Solve the compound inequality. Then graph the solution set.
Solve both inequalities for y.
The solution set is all points that satisfy
{y < –4 or y ≥ –2}.
6y < –24 OR y +5 ≥ 3
6y < –24
–6 –5 –4 –3 –2 –1 0 1 2 3
Holt Algebra 2
Solving 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.
[–4, 0)
Solve the compound inequality. Then graph the solution set.
c ≥ –4
c < 0
–6 –5 –4 –3 –2 –1 0 1 2 3
and
Solve the compound inequality. Then graph the solution set.
Solve both inequalities for x.
The solution set is the set of all points that satisfy
{x < 3 or x ≥ 5}.
x – 5 < –2 OR –2x ≤ –10
x < 3 x ≥ 5
(–∞, 3) U [5, ∞)
–3 –2 –1 0 1 2 3 4 5 6
Holt Algebra 2
x ≥ –3 x < 4
The solution set is the set of points that satisfy both {x ≥ –3 x
< 4}.
[–3, 4)
Check It Out! Example 1b
–4 –3 –2 –1 0 1 2 3 4 5
Holt Algebra 2
Solving 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
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
Solving 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
Solving 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
Think: Greator inequalities involving > or ≥ symbols are
disjunctions.
Think: Less thand inequalities involving < or
≤ symbols are conjunctions.
Solving 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
Solve the equation.
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
k = 13 or k = –7
Holt Algebra 2
|x + 9| = 13
Solve the equation.
Rewrite the absolute value as a disjunction.
This can be read as “the distance from x to +9 is 4.”
Subtract 9 from both sides of each equation.
x + 9 = 13 or x + 9 = –13
x = 4 or x = –22
Holt Algebra 2
|6x| – 8 = 22
Solve the equation.
Divide both sides of each equation by 6.
|6x| = 30
Holt Algebra 2
Solving Absolute-Value Equations and Inequalities
You can solve absolute-value inequalities using the same methods
that are used to solve an absolute-value equation.
Holt Algebra 2
Example 3A: Solving Absolute-Value Inequalities with
Disjunctions
Solve the inequality. Then graph the solution.
Rewrite the absolute value as a disjunction.
|–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
Example 3A Continued
(–∞, –2] U [3, ∞)
To check, you can test a point in each of the three region.
|–4(–3) + 2| ≥ 10
–3 –2 –1 0 1 2 3 4 5 6
Holt Algebra 2
Solve the inequality. Then graph the solution.
Rewrite the absolute value as a disjunction.
|4x – 8| > 12
4x > 20 or 4x < –4
Divide both sides of each inequality by 4.
Add 8 to both sides of each inequality.
x > 5 or x < –1
Check It Out! Example 3a
Holt Algebra 2
(–∞, –1) U (5, ∞)
To check, you can test a point in each of the three region.
|4(–2) + 8| > 12
–3 –2 –1 0 1 2 3 4 5 6
Holt Algebra 2
Solve the inequality. Then graph the solution.
|3x| + 36 > 12
Isolate the absolute value as a disjunction.
Rewrite the absolute value as a disjunction.
Check It Out! Example 3b
3x > –24 or 3x < 24
x > –8 or x < 8
|3x| > –24
–3 –2 –1 0 1 2 3 4 5 6
Holt Algebra 2
Solve the compound inequality. Then graph the solution set.
Example 4A: Solving Absolute-Value Inequalities with
Conjunctions
|2x +7| ≤ 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
Example 4A Continued
–6 –5 –3 –2 –1 0 1 2 3 4
Holt Algebra 2
Solve the compound inequality. Then graph the solution set.
|x – 5| ≤ 8
Add 5 to both sides of each inequality.
Rewrite the absolute value as a conjunction.
x – 5 ≤ 8 and x – 5 ≥ –8
x ≤ 13 and x ≥ –3
Check It Out! Example 4a
Holt Algebra 2
The solution set is {–3 ≤ x ≤ 13}.
Check It Out! Example 4
–10 –5 0 5 10 15 20 25
Holt Algebra 2
Solve the compound inequality. Then graph the solution set.
–2|x +5| > 10
Divide both sides by –2, and reverse the inequality symbol.
Subtract 5 from both sides of each inequality.
Rewrite the absolute value as a conjunction.
x + 5 < –5 and x + 5 > 5
x < –10 and x > 0
Because no real number satisfies both x < –10 and x > 0,
there is no solution. The solution set is ø.
Check It Out! Example 4b
|x + 5| < –5