Solving Compound Inequalities
LESSON
3-6
Solving Compound Inequalities
Additional Examples Example 1 The pH level of a popular shampoo is between 6.0 and 6.5 inclusive. Write a compound inequality to show the pH levels of this shampoo. Graph the solutions. Let p be 6.0
is less than or equal to
. pH level
is less than or equal to
6.0
6.5 6.5
6.0 ! p ! 6.5
Example 2 Solve each compound inequality and graph the solutions. A.
!5 ! x " 1 ! 2 –5 " x # 1 " 2
Since 1 is added to x, from each part of the inequality. The solution set is . Graph Graph
. .
Graph the by finding where the two graphs .
Copyright © by Holt, Rinehart and Winston. All rights reserved.
102
Algebra 1
LESSON
3-6
Solving Compound Inequalities
Additional Examples Example 1 The pH level of a popular shampoo is between 6.0 and 6.5 inclusive. Write a compound inequality to show the pH levels of this shampoo. Graph the solutions. Let p be the pH level of the shampoo.
.
6.0
is less than or equal to
pH level
is less than or equal to
6.5
6.0
!
p
!
6.5
6.0 ! p ! 6.5 6.0 6.1 6.2 6.3 6.4 6.5
Example 2 Solve each compound inequality and graph the solutions. A.
!5 " x " 1 " 2 –5 " x # 1 " 2 !1
!1
!6 " x " 1
!1
Since 1 is added to x, subtract 1 from each part of the inequality. The solution set is {x : !6 " x AND x " 1
–6 –5 –4 –3 –2 –1 0 1
Graph !6 " x
–6 –5 –4 –3 –2 –1 0 1
Graph x " 1
–6 –5 –4 –3 –2 –1 0 1
. .
Graph the intersection by finding where the two graphs overlap
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.
102
.
Algebra 1
LESSON 3-6 CONTINUED
B.
8 ! 3x " 1 # 11
9
8 ! 3x " 1 # 11
Since 1 is subtracted from 3x, to each part of the inequality.
9 ! 3x # 12
Since x is multiplied by 3,
!
3x
#
12 to undo the
.
The solution set is .
Graph
.
Graph
.
Graph the finding where the two graphs
Copyright © by Holt, Rinehart and Winston. All rights reserved.
103
by .
Algebra 1
LESSON 3-6 CONTINUED
B.
8 ! 3x " 1 # 11 8 ! 3x " 1 # 11
!1
!1
!1 Since x is multiplied by 3, divide
9 ! 3x # 12 9 3
!
3x
Since 1 is subtracted from 3x, add 1 to each part of the inequality.
#
3
12 3
each part of the inequality by 3 to undo the multiplication
.
The solution set is 3!x#4
{x : 3 ! x AND x # 4}
2 2.5 3 3.5 4 4.5 5
Graph 3 ! x .
2 2.5 3 3.5 4 4.5 5
Graph x # 4 .
2 2.5 3 3.5 4 4.5 5
Graph the intersection
.
by
finding where the two graphs overlap
Copyright © by Holt, Rinehart and Winston. All rights reserved.
103
.
Algebra 1
LESSON 3-6 CONTINUED
Example 3 Solve each compound inequality and graph the solutions. A.
8 ! t " 7 OR 8 ! t # 2 8!t"7
OR 8 ! t # 2 Solve each simple inequality. The solution set is . Graph
.
Graph
.
Graph the
by the regions.
B.
4x $ 20 OR 3x % 21 4x $ 20 OR 3x % 21 4x
$
20
OR
3x
%
21 Solve each simple inequality. The solution set is . Graph
.
Graph
.
Graph the
by the regions.
Copyright © by Holt, Rinehart and Winston. All rights reserved.
104
Algebra 1
LESSON 3-6 CONTINUED
Example 3 Solve each compound inequality and graph the solutions. A.
8 ! t " 7 OR 8 ! t # 2 8!t"7 !8
OR 8 ! t # 2
!8
!8
!8
t " !1 OR t # !6
Solve each simple inequality. The solution set is {t : t " !1 OR t # !6}
–7 –6 –5 –4 –3 –2 –1
0
Graph t " !1
.
–7 –6 –5 –4 –3 –2 –1
0
Graph t # !6
.
–7 –6 –5 –4 –3 –2 –1
0
Graph the union combining
B.
.
by the regions.
4x $ 20 OR 3x % 21 4x $ 20 OR 3x % 21 4x
$
20
4
OR
4
3x 3
x $ 5 OR x % 7
%
21 3
Solve each simple inequality. The solution set is {x : x $ 5 OR x % 7}
2
3
4
5
6
7
8
Graph x $ 5 .
2
3
4
5
6
7
8
Graph x % 7 .
2
3
4
5
6
7
8
Graph the union combining
Copyright © by Holt, Rinehart and Winston. All rights reserved.
104
.
by the regions.
Algebra 1
LESSON 3-6 CONTINUED
Example 4 Write the compound inequality shown by each graph. A.
–10 –8 –6 –4 –2
0
2
The shaded portion of the graph is , so the compound involves
.
On the left, the graph shows an arrow pointing or
. The
, so use either
circle at !8 means !8 is so use
.
x" On the right, the graph shows an arrow pointing either
or
. The
, so use
circle at 0 means 0 is , so use
.
x# The compound inequality is
Copyright © by Holt, Rinehart and Winston. All rights reserved.
.
105
Algebra 1
LESSON 3-6 CONTINUED
Example 4 Write the compound inequality shown by each graph. A.
–10 –8 –6 –4 –2
0
2
The shaded portion of the graph is not between two values
, so the compound involves OR
.
On the left, the graph shows an arrow pointing left %
or ! . The solid
a solution
, so use either
circle at !8 means !8 is
so use ! .
x " $8 On the right, the graph shows an arrow pointing right either "
or # . The empty
not a solution
, so use
circle at 0 means 0 is
, so use " .
x# 0 The compound inequality is x ! !8 OR x " 0
Copyright © by Holt, Rinehart and Winston. All rights reserved.
105
.
Algebra 1
LESSON 3-6 CONTINUED
B.
5 –4 –2
0
2
4
6
8
The shaded portion of the graph is , so the compound involves
.
The shaded values are on the or
. The
, so use either circle at –2 means –2 is so use
.
m! The shaded values are to the The use
, so use either " or
circle at 5 means 5 is
. , so
.
m" The compound inequality is
Copyright © by Holt, Rinehart and Winston. All rights reserved.
.
106
Algebra 1
LESSON 3-6 CONTINUED
B.
5 –4 –2
0
2
4
6
8
The shaded portion of the graph is between the values !2 and 5
, so the compound involves AND
The shaded values are on the right of !2 !
or $ . The empty
not a solution
. , so use either
circle at –2 means –2 is so use ! .
m ! !2 The shaded values are to the left of 5 The empty
, so use either " or # .
circle at 5 means 5 is not a solution
, so
use " . m" 5 The compound inequality is m ! !2 AND m " 5 (or !2 " m " 5)
Copyright © by Holt, Rinehart and Winston. All rights reserved.
106
.
Algebra 1
LESSON 3-6 CONTINUED
Check It Out! 1. The free chlorine level in a pool should be between 1.0 and 3.0 parts per million inclusive. Write a compound inequality to show the levels that are within this range. Graph the solutions.
2. Solve the compound inequality and graph the solutions. !4 " 3n # 5 $ 11
3. Solve the compound inequality and graph the solutions. 2 # r $ 12 OR r # 5 % 19
4. Write the compound inequality shown by the graph. –5 –4 –3 –2 –1
0
1
Copyright © by Holt, Rinehart and Winston. All rights reserved.
2
3
4
5
107
Algebra 1
LESSON 3-6 CONTINUED
Check It Out! 1. The free chlorine level in a pool should be between 1.0 and 3.0 parts per million inclusive. Write a compound inequality to show the levels that are within this range. Graph the solutions.
1.0 ! c ! 3.0;
0
1
2
3
4
5
6
2. Solve the compound inequality and graph the solutions. !4 " 3n # 5 $ 11
%3 ! n # 2;
–3 –2 –1
0
1
2
3
3. Solve the compound inequality and graph the solutions. 2 # r $ 12 OR r # 5 % 19
r # 10 OR r $ 14;
4
6
8 10 12 14 16
4. Write the compound inequality shown by the graph. –5 –4 –3 –2 –1
0
1
2
3
4
5
x ! !3 OR x " 2
Copyright © by Holt, Rinehart and Winston. All rights reserved.
107
Algebra 1
3-6
Solving Compound Inequalities
Additional Examples Example 1 The pH level of a popular shampoo is between 6.0 and 6.5 inclusive. Write a compound inequality to show the pH levels of this shampoo. Graph the solutions. Let p be 6.0
is less than or equal to
. pH level
is less than or equal to
6.0
6.5 6.5
6.0 ! p ! 6.5
Example 2 Solve each compound inequality and graph the solutions. A.
!5 ! x " 1 ! 2 –5 " x # 1 " 2
Since 1 is added to x, from each part of the inequality. The solution set is . Graph Graph
. .
Graph the by finding where the two graphs .
Copyright © by Holt, Rinehart and Winston. All rights reserved.
102
Algebra 1
LESSON
3-6
Solving Compound Inequalities
Additional Examples Example 1 The pH level of a popular shampoo is between 6.0 and 6.5 inclusive. Write a compound inequality to show the pH levels of this shampoo. Graph the solutions. Let p be the pH level of the shampoo.
.
6.0
is less than or equal to
pH level
is less than or equal to
6.5
6.0
!
p
!
6.5
6.0 ! p ! 6.5 6.0 6.1 6.2 6.3 6.4 6.5
Example 2 Solve each compound inequality and graph the solutions. A.
!5 " x " 1 " 2 –5 " x # 1 " 2 !1
!1
!6 " x " 1
!1
Since 1 is added to x, subtract 1 from each part of the inequality. The solution set is {x : !6 " x AND x " 1
–6 –5 –4 –3 –2 –1 0 1
Graph !6 " x
–6 –5 –4 –3 –2 –1 0 1
Graph x " 1
–6 –5 –4 –3 –2 –1 0 1
. .
Graph the intersection by finding where the two graphs overlap
Copyright © by Holt, Rinehart and Winston. All rights reserved.
.
102
.
Algebra 1
LESSON 3-6 CONTINUED
B.
8 ! 3x " 1 # 11
9
8 ! 3x " 1 # 11
Since 1 is subtracted from 3x, to each part of the inequality.
9 ! 3x # 12
Since x is multiplied by 3,
!
3x
#
12 to undo the
.
The solution set is .
Graph
.
Graph
.
Graph the finding where the two graphs
Copyright © by Holt, Rinehart and Winston. All rights reserved.
103
by .
Algebra 1
LESSON 3-6 CONTINUED
B.
8 ! 3x " 1 # 11 8 ! 3x " 1 # 11
!1
!1
!1 Since x is multiplied by 3, divide
9 ! 3x # 12 9 3
!
3x
Since 1 is subtracted from 3x, add 1 to each part of the inequality.
#
3
12 3
each part of the inequality by 3 to undo the multiplication
.
The solution set is 3!x#4
{x : 3 ! x AND x # 4}
2 2.5 3 3.5 4 4.5 5
Graph 3 ! x .
2 2.5 3 3.5 4 4.5 5
Graph x # 4 .
2 2.5 3 3.5 4 4.5 5
Graph the intersection
.
by
finding where the two graphs overlap
Copyright © by Holt, Rinehart and Winston. All rights reserved.
103
.
Algebra 1
LESSON 3-6 CONTINUED
Example 3 Solve each compound inequality and graph the solutions. A.
8 ! t " 7 OR 8 ! t # 2 8!t"7
OR 8 ! t # 2 Solve each simple inequality. The solution set is . Graph
.
Graph
.
Graph the
by the regions.
B.
4x $ 20 OR 3x % 21 4x $ 20 OR 3x % 21 4x
$
20
OR
3x
%
21 Solve each simple inequality. The solution set is . Graph
.
Graph
.
Graph the
by the regions.
Copyright © by Holt, Rinehart and Winston. All rights reserved.
104
Algebra 1
LESSON 3-6 CONTINUED
Example 3 Solve each compound inequality and graph the solutions. A.
8 ! t " 7 OR 8 ! t # 2 8!t"7 !8
OR 8 ! t # 2
!8
!8
!8
t " !1 OR t # !6
Solve each simple inequality. The solution set is {t : t " !1 OR t # !6}
–7 –6 –5 –4 –3 –2 –1
0
Graph t " !1
.
–7 –6 –5 –4 –3 –2 –1
0
Graph t # !6
.
–7 –6 –5 –4 –3 –2 –1
0
Graph the union combining
B.
.
by the regions.
4x $ 20 OR 3x % 21 4x $ 20 OR 3x % 21 4x
$
20
4
OR
4
3x 3
x $ 5 OR x % 7
%
21 3
Solve each simple inequality. The solution set is {x : x $ 5 OR x % 7}
2
3
4
5
6
7
8
Graph x $ 5 .
2
3
4
5
6
7
8
Graph x % 7 .
2
3
4
5
6
7
8
Graph the union combining
Copyright © by Holt, Rinehart and Winston. All rights reserved.
104
.
by the regions.
Algebra 1
LESSON 3-6 CONTINUED
Example 4 Write the compound inequality shown by each graph. A.
–10 –8 –6 –4 –2
0
2
The shaded portion of the graph is , so the compound involves
.
On the left, the graph shows an arrow pointing or
. The
, so use either
circle at !8 means !8 is so use
.
x" On the right, the graph shows an arrow pointing either
or
. The
, so use
circle at 0 means 0 is , so use
.
x# The compound inequality is
Copyright © by Holt, Rinehart and Winston. All rights reserved.
.
105
Algebra 1
LESSON 3-6 CONTINUED
Example 4 Write the compound inequality shown by each graph. A.
–10 –8 –6 –4 –2
0
2
The shaded portion of the graph is not between two values
, so the compound involves OR
.
On the left, the graph shows an arrow pointing left %
or ! . The solid
a solution
, so use either
circle at !8 means !8 is
so use ! .
x " $8 On the right, the graph shows an arrow pointing right either "
or # . The empty
not a solution
, so use
circle at 0 means 0 is
, so use " .
x# 0 The compound inequality is x ! !8 OR x " 0
Copyright © by Holt, Rinehart and Winston. All rights reserved.
105
.
Algebra 1
LESSON 3-6 CONTINUED
B.
5 –4 –2
0
2
4
6
8
The shaded portion of the graph is , so the compound involves
.
The shaded values are on the or
. The
, so use either circle at –2 means –2 is so use
.
m! The shaded values are to the The use
, so use either " or
circle at 5 means 5 is
. , so
.
m" The compound inequality is
Copyright © by Holt, Rinehart and Winston. All rights reserved.
.
106
Algebra 1
LESSON 3-6 CONTINUED
B.
5 –4 –2
0
2
4
6
8
The shaded portion of the graph is between the values !2 and 5
, so the compound involves AND
The shaded values are on the right of !2 !
or $ . The empty
not a solution
. , so use either
circle at –2 means –2 is so use ! .
m ! !2 The shaded values are to the left of 5 The empty
, so use either " or # .
circle at 5 means 5 is not a solution
, so
use " . m" 5 The compound inequality is m ! !2 AND m " 5 (or !2 " m " 5)
Copyright © by Holt, Rinehart and Winston. All rights reserved.
106
.
Algebra 1
LESSON 3-6 CONTINUED
Check It Out! 1. The free chlorine level in a pool should be between 1.0 and 3.0 parts per million inclusive. Write a compound inequality to show the levels that are within this range. Graph the solutions.
2. Solve the compound inequality and graph the solutions. !4 " 3n # 5 $ 11
3. Solve the compound inequality and graph the solutions. 2 # r $ 12 OR r # 5 % 19
4. Write the compound inequality shown by the graph. –5 –4 –3 –2 –1
0
1
Copyright © by Holt, Rinehart and Winston. All rights reserved.
2
3
4
5
107
Algebra 1
LESSON 3-6 CONTINUED
Check It Out! 1. The free chlorine level in a pool should be between 1.0 and 3.0 parts per million inclusive. Write a compound inequality to show the levels that are within this range. Graph the solutions.
1.0 ! c ! 3.0;
0
1
2
3
4
5
6
2. Solve the compound inequality and graph the solutions. !4 " 3n # 5 $ 11
%3 ! n # 2;
–3 –2 –1
0
1
2
3
3. Solve the compound inequality and graph the solutions. 2 # r $ 12 OR r # 5 % 19
r # 10 OR r $ 14;
4
6
8 10 12 14 16
4. Write the compound inequality shown by the graph. –5 –4 –3 –2 –1
0
1
2
3
4
5
x ! !3 OR x " 2
Copyright © by Holt, Rinehart and Winston. All rights reserved.
107
Algebra 1
