holt algebra 2 2-8 solving absolute-value equations and inequalities a compound statement is made up...

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


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


Dis- means “apart.” Disjunctions have two separate pieces. Con- means “together” Conjunctions represent one piece.

Reading Math

Holt Algebra 2


Example 1: 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


Solve both inequalities for x.

–3x < –12 and x + 4 ≤ 12

2 3 4 5 6 7 8 9 10 11

–3x < –12 AND x + 4 ≤ 12

x > 4 x ≤ 8

The solution set is the set of points that satisfy both {x|4 < x ≤ 8}.

(4, 8]


Example 2

Holt Algebra 2


Example 3: Solving Compound Inequalities


Solve both inequalities for x.

The solution set is the set of all points that satisfy {x|x < 4 or x > 6}.

–3 –2 –1 0 1 2 3 4 5 6

X +3 < 7 OR 3x > 18

x < 4 x > 6

x + 3 < 7 or 3x > 18

(–∞, 4) U (6, ∞)

Holt Algebra 2


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


Solve the equation.

Example 2A: 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


Check It Out! Example 2b

|6x| – 8 = 22

Solve the equation.

Isolate the absolute-value expression.

Rewrite the absolute value as a disjunction. Divide both sides of each equation by 6.

|6x| = 30

6x = 30 or 6x = –30

x = 5 or x = –5

Holt Algebra 2


You can solve absolute-value inequalities using the same methods that are used to solve an absolute-value equation.

Holt Algebra 2


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

Holt Algebra 2


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



|x – 5| ≤ 8 Multiply both sides by 2.

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 {x|–3 ≤ x ≤ 13}.

Check It Out! Example 4

–10 –5 0 5 10 15 20 25

Holt Algebra 2


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


Example 3A: Solving Absolute-Value Inequalities with Disjunctions

Solve the inequality. Then graph the solution.


|–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


Example 3A Continued

{q|q ≤ –2 or q ≥ 3}

–3 –2 –1 0 1 2 3 4 5 6

(–∞, –2] U [3, ∞)

To check, you can test a point in each of the three region.

|–4(–3) + 2| ≥ 10

|14| ≥ 10

|–4(0) + 2| ≥ 10

|2| ≥ 10 x

|–4(4) + 2| ≥ 10

|–14| ≥ 10

Holt Algebra 2




|4x – 8| > 12

4x – 8 > 12 or 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


{x|x < –1 or x > 5}

–3 –2 –1 0 1 2 3 4 5 6

(–∞, –1) U (5, ∞)

To check, you can test a point in each of the three region.

|4(–2) + 8| > 12

|–16| > 12

|4(0) + 8| > 12

|8| > 12 x

|4(6) + 8| > 12

|32| > 12

Check It Out! Example 3a Continued

Holt Algebra 2



|3x| + 36 > 12

Divide both sides of each inequality by 3.

Isolate the absolute value as a disjunction.



3x > –24 or 3x < 24

x > –8 or x < 8

|3x| > –24

–3 –2 –1 0 1 2 3 4 5 6

(–∞, ∞)

The solution is all real numbers, R.

Holt Algebra 2



–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.


|x + 5| < –5

holt algebra 2 2-8 solving absolute-value equations and inequalities a compound statement is made up...

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