Slide 1
2-8Solving Absolute-Value Equations and InequalitiesHolt Algebra
2Warm UpLesson PresentationLesson QuizHolt Algebra 22-8Solving
Absolute-Value Equations and Inequalities Warm Up
Solve.1. y + 7 < 112. 4m 123. 5 2x 17y < 18m 3x 6Use
interval notation to indicate the graphed numbers.4.5.
(-2, 3](-, 1]Holt Algebra 22-8Solving Absolute-Value Equations
and Inequalities Solve compound inequalities.
Write and solve absolute-value equations and inequalities.
ObjectivesHolt Algebra 22-8Solving Absolute-Value Equations and
Inequalities disjunctionconjunctionabsolute-valueVocabularyHolt
Algebra 22-8Solving 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 22-8Solving 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
UHolt Algebra 22-8Solving Absolute-Value Equations and
Inequalities Dis- means apart. Disjunctions have two separate
pieces. Con- means together Conjunctions represent one
piece.Reading MathHolt Algebra 22-8Solving Absolute-Value Equations
and Inequalities Example 1A: Solving Compound InequalitiesSolve 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 36y < 24y + 5 3y < 4y
2or 6 5 4 3 2 1 0 1 2 3(, 4) U [2, )Holt Algebra 22-8Solving
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 4c < 0 6 5 4 3 2 1 0 1 2 3[4, 0)Example 1B:
Solving Compound InequalitiesSolve the compound inequality. Then
graph the solution set.
and
2c + 1 < 1Holt Algebra 22-8Solving Absolute-Value Equations
and Inequalities Example 1C: Solving Compound InequalitiesSolve 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|x < 3 or x 5}. 3 2 1 0 1 2 3 4 5 6x 5 < 2 OR 2x 10
x < 3 x 5 x 5 < 2 or 2x 10(, 3) U [5, )Holt Algebra
22-8Solving Absolute-Value Equations and Inequalities Solve the
compound inequality. Then graph the solution set.Solve both
inequalities for x. The solution set is all points that satisfy
{x|x < 3 U x 6}.x 2 < 1 OR 5x 30x 2 < 15x 30x < 3x 6or
1 0 1 2 3 4 5 6 7 8(, 3) U [6, )Check It Out! Example 1a Holt
Algebra 22-8Solving Absolute-Value Equations and Inequalities Solve
both inequalities for x. 2x 6 and x > 4 4 3 2 1 0 1 2 3 4 52x 6
AND x > 4x 3 x < 4The solution set is the set of points that
satisfy both {x|x 3 x < 4}. U [3, 4) Solve the compound
inequality. Then graph the solution set.Check It Out! Example 1b
Holt Algebra 22-8Solving Absolute-Value Equations and Inequalities
Solve both inequalities for x. x 5 < 12 or 6x 12 2 4 6 8 10 12
14 16 18 20x 5 < 12 OR 6x 12x < 17 x 2Because every point
that satisfies x < 2 also satisfies x < 2, the solution set
is {x|x < 17}. (-, 17) Solve the compound inequality. Then graph
the solution set.Check It Out! Example 1c Holt Algebra 22-8Solving
Absolute-Value Equations and Inequalities 13Solve both inequalities
for x. 3x < 12 and x + 4 12 2 3 4 5 6 7 8 9 10 113x < 12 AND
x + 4 12x < 4 x 8The solution set is the set of points that
satisfy both {x|4 < x 8}. (4, 8] Solve the compound inequality.
Then graph the solution set.Check It Out! Example 1d Holt Algebra
22-8Solving 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 22-8Solving 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 22-8Solving 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 22-8Solving 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 22-8Solving Absolute-Value Equations and
Inequalities Think: Greator inequalities involving > or symbols
are disjunctions. Think: Less thand inequalities involving < or
symbols are conjunctions.Helpful HintHolt Algebra 22-8Solving
Absolute-Value Equations and Inequalities Note: The symbol can
replace , and the rules still apply.
Holt Algebra 22-8Solving Absolute-Value Equations and
Inequalities Solve the equation.Example 2A: Solving Absolute-Value
EquationsRewrite 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| = 103 + k = 10 or 3 + k = 10k = 13 or k = 7Holt
Algebra 22-8Solving Absolute-Value Equations and Inequalities Solve
the equation.Example 2B: Solving Absolute-Value Equationsx = 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 22-8Solving Absolute-Value Equations and
Inequalities Check It Out! Example 2a |x + 9| = 13Solve 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 = 22Holt
Algebra 22-8Solving Absolute-Value Equations and Inequalities Check
It Out! Example 2b |6x| 8 = 22Solve the equation.Isolate the
absolute-value expression. Rewrite the absolute value as a
disjunction. Divide both sides of each equation by 6. |6x| = 306x =
30 or 6x = 30x = 5 or x = 5Holt Algebra 22-8Solving 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 22-8Solving Absolute-Value Equations and
Inequalities Example 3A: Solving Absolute-Value Inequalities with
Disjunctions Solve the inequality. Then graph the solution.Rewrite
the absolute value as a disjunction.|4q + 2| 104q + 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 22-8Solving Absolute-Value
Equations and Inequalities 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 22-8Solving Absolute-Value Equations
and Inequalities Solve the inequality. Then graph the
solution.|0.5r| 3 30.5r 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| 0Rewrite the absolute value as a disjunction.Example
3B: 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
22-8Solving Absolute-Value Equations and Inequalities Solve the
inequality. Then graph the solution.Rewrite the absolute value as a
disjunction.|4x 8| > 124x 8 > 12 or 4x 8 < 124x > 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 22-8Solving Absolute-Value Equations and
Inequalities {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 ContinuedHolt
Algebra 22-8Solving Absolute-Value Equations and Inequalities Solve
the inequality. Then graph the solution.|3x| + 36 > 12Divide
both sides of each inequality by 3.Isolate the absolute value as a
disjunction.Rewrite the absolute value as a disjunction.Check It
Out! Example 3b 3x > 24 or 3x < 24x > 8 or x < 8|3x|
> 243 2 1 0 1 2 3 4 5 6(, )The solution is all real numbers,
R.Holt Algebra 22-8Solving Absolute-Value Equations and
Inequalities Solve the compound inequality. Then graph the solution
set.Example 4A: Solving Absolute-Value Inequalities with
Conjunctions|2x +7| 3Multiply 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 22-8Solving Absolute-Value Equations and
Inequalities Example 4A ContinuedThe solution set is {x|5 x 2}.6 5
3 2 1 0 1 2 3 4Holt Algebra 22-8Solving Absolute-Value Equations
and Inequalities Solve the compound inequality. Then graph the
solution set.Example 4B: 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 .
Holt Algebra 22-8Solving Absolute-Value Equations and
Inequalities Solve the compound inequality. Then graph the solution
set.|x 5| 8Multiply 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 22-8Solving Absolute-Value Equations and
Inequalities The solution set is {x|3 x 13}.Check It Out! Example 4
10 5 0 5 10 15 20 25Holt Algebra 22-8Solving Absolute-Value
Equations and Inequalities Solve the compound inequality. Then
graph the solution set.2|x +5| > 10Divide 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 .Check It Out! Example 4b |x + 5| < 5Holt
Algebra 22-8Solving Absolute-Value Equations and Inequalities
Lesson Quiz: Part IISolve. Then graph the solution.5.6.7.|1 2x|
> 7{x|x < 3 or x > 4}|3k| + 11 > 8R2|u + 7| 164 3 2 1 0
1 2 3 4 54 3 2 1 0 1 2 3 4 5Holt Algebra 22-8Solving Absolute-Value
Equations and Inequalities 39
