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Warm Up
Compare. Write <, >, or =.
1. −3 < 2
2. 6.5 > 6.3
3.
4. 0.25 =
>
Tell whether the inequality x < 5 is true
or false for the following values of x.
5. x = –10
T
6. x = 5
7. x = 4.99
T
8. x =
F
T
An inequality is a statement that two quantities
are not equal. The quantities are compared by
using the following signs:
≥
≠
A≤B
A≥B
A≠B
A is less
than or
equal to B.
A is greater
than or
equal to B.
A is not
equal to B.
<
>
≤
A<B
A>B
A is less
than B.
A is greater
than B.
A solution of an inequality is any value that
makes the inequality true.
An inequality like 3 + x < 9
has too many solutions to
list. You can use a graph on
a number line to show all
the solutions.
The solutions are shaded and an arrow shows that
the solutions continue past those shown on the
graph. To show that an endpoint is a solution, draw a
solid circle at the number. To show an endpoint is
not a solution, draw an empty circle.
Example 2: Graphing Inequalities
Graph each inequality.
A. m ≥
–
0
Draw a solid circle at
1
2
3
3
B. t < 5(–1 + 3)
t < 5(–1 + 3)
t < 5(2)
t < 10
–8 –6 –4 –2 0
2
4
6
8
10 12
.
Shade all the numbers
greater than and draw an
arrow pointing to the right.
Simplify.
Draw an empty circle at
10.
Shade all the numbers
less than 10 and draw an
arrow pointing to the left.
Check It Out! Example 2
Graph each inequality.
Draw an empty circle at 2.5.
a. c > 2.5
2.5
–4 –3 –2 –1
0
1
2
3
4
5
6
b. 22 – 4 ≥ w
22 – 4 ≥ w
4–4≥
w 0≥w
–4 –3 –2 –1 0
1
Draw a solid circle at 0.
Shade in all numbers less than 0 and
draw an arrow pointing to the left.
2
3
4
5
6
c. m ≤ –3
Draw a solid circle at –3.
−3
–8 –6 –4 –2
0
Shade in all the numbers greater
than 2.5 and draw an arrow pointing
to the right.
2
4
6
8
10 12
Shade in all numbers less than –3
and draw an arrow pointing to the left.
Example 3: Writing an Inequality from a Graph
Write the inequality shown by each graph.
x<2
Use any variable. The arrow points to the left, so use
either < or ≤. The empty circle at 2 means that 2 is
not a solution, so use <.
x ≥ –0.5
Use any variable. The arrow points to the right, so
use either > or ≥. The solid circle at –0.5 means
that –0.5 is a solution, so use ≥.
Check It Out! Example 3
Write the inequality shown by the graph.
x < 2.5
Use any variable. The arrow
points to the left, so use either <
or ≤. The empty circle at 2.5
means that 2.5 is not a solution,
so use so use <.
Reading Math
“No more than” means “less than or
equal to.”
“At least” means “greater than or
equal to”.
Example 4: Application
Ray’s dad told him not to turn on the air
conditioner unless the temperature is at least
85°F. Define a variable and write an inequality
for the temperatures at which Ray can turn on
the air conditioner. Graph the solutions.
Let t represent the temperatures at which Ray can
turn on the air conditioner.
Turn on the AC when temperature
t
≥
t  85
70
75
80
85
is at least 85°F
90
85
Draw a solid circle at 85. Shade
all numbers greater than 85 and
draw an arrow pointing to the
right.
Check It Out! Example 4
A store’s employees earn at least $8.50 per
hour. Define a variable and write an
inequality for the amount the employees
may earn per hour. Graph the solutions.
Let w represent an employee’s wages.
An employee earns
at least
w
≥
w ≥ 8.5
−2 0
2 4
8.5
6
8 10 12 14 16 18
$8.50
8.50
Lesson Quiz: Part I
1. Describe the solutions of 7 < x + 4.
all real numbers greater than 3
2. Graph h ≥ –4.75
–5
–4.75
–4.5
Write the inequality shown by each graph.
3.
4.
x≥3
x < –5.5
Lesson Quiz: Part II
5. A cell phone plan offers free minutes for no more
than 250 minutes per month. Define a variable
and write an inequality for the possible number of
free minutes. Graph the solution.
Let m = number of minutes
0 ≤ m ≤ 250
0
250
Solving one-step inequalities is much like solving onestep equations. To solve an inequality, you need to isolate
the variable using the properties of inequality and inverse
operations.
Example 1A: Using Addition and Subtraction to Solve
Inequalities
Solve the inequality and graph the solutions.
x + 12 < 20
x + 12 < 20
Since 12 is added to x, subtract
12 from both sides to undo the
addition.
–12 –12
x+0 < 8
x < 8
Draw an empty circle at 8.
–10 –8 –6 –4 –2
0
2
4
6
8 10
Shade all numbers less than 8
and draw an arrow pointing to
the left.
Example 1B: Using Addition and Subtraction to Solve
Inequalities
Solve the inequality and graph the solutions.
d – 5 > –7
d – 5 > –7
+5
Since 5 is subtracted from d, add
5 to both sides to undo the
subtraction.
+5
d + 0 > –2
d > –2
Draw an empty circle at –2.
–10 –8 –6 –4 –2
0
2
4
6
8 10
Shade all numbers greater than –2
and draw an arrow pointing to the
right.
Example 1C: Using Addition and Subtraction to Solve
Inequalities
Solve the inequality and graph the solutions.
0.9 ≥ n – 0.3
0.9 ≥ n – 0.3
+0.3
Since 0.3 is subtracted from n, add
0.3 to both sides to undo the
subtraction.
+0.3
1.2 ≥ n – 0
1.2 ≥ n
0
1
1.2

Draw a solid circle at 1.2.
2
Shade all numbers less than 1.2
and draw an arrow pointing to
the left.
Check It Out! Example 1
Solve each inequality and graph the solutions.
a. s + 1 ≤ 10
s + 1 ≤ 10
–1
Since 1 is added to s, subtract 1 from both sides
to undo the addition.
–1
s+0≤ 9
9
–10 –8 –6 –4 –2
0
2
4
6
8 10
s ≤ 9
b.
> –3 + t
> –3 + t
+3
Since –3 is added to t, add 3 to both sides to
undo the addition.
+3
> 0+t
–10 –8 –6 –4 –2
t<
0
2
4
6
8 10
Check It Out! Example 1c
Solve the inequality and graph the solutions.
q – 3.5 < 7.5
q – 3.5 < 7.5
+ 3.5 +3.5
Since 3.5 is subtracted from q, add 3.5 to
both sides to undo the subtraction.
q – 0 < 11
q < 11
–7 –5 –3 –1
1
3
5
7
9 11 13
Since there can be an infinite number of solutions to an
inequality, it is not possible to check all the solutions. You can
check the endpoint and the direction of the inequality symbol.
The solutions of x + 9 < 15 are given by x < 6.
Lesson Quiz: Part I
Solve each inequality and graph the solutions.
1. 13 < x + 7
x>6
2. –6 + h ≥ 15
h ≥ 21
3. 6.7 + y ≤ –2.1
y ≤ –8.8
Lesson Quiz: Part II
4. A certain restaurant has room for 120 customers. On
one night, there are 72 customers dining. Write and
solve an inequality to show how many more people
can eat at the restaurant.
x + 72 ≤ 120; x ≤ 48, where x is a natural
number
Example 1A: Multiplying or Dividing by a Positive
Number
Solve the inequality and graph the solutions.
7x > –42
7x > –42
>
Since x is multiplied by 7, divide both
sides by 7 to undo the multiplication.
1x > –6
x > –6
–10 –8 –6 –4 –2 0
2 4
6 8 10
Example 1B: Multiplying or Dividing by a Positive
Number
Solve the inequality and graph the solutions.
3(2.4) ≤ 3
Since m is divided by 3, multiply both
sides by 3 to undo the division.
7.2 ≤ m(or m ≥ 7.2)
0 2 4
6 8 10 12 14 16 18 20
Example 1C: Multiplying or Dividing by a Positive
Number
Solve the inequality and graph the solutions.
r < 16
0 2 4
Since r is multiplied by ,
multiply both sides by the
reciprocal of .
6 8 10 12 14 16 18 20
Check It Out! Example 1a
Solve the inequality and graph the solutions.
4k > 24
Since k is multiplied by 4, divide
both sides by 4.
k>6
0 2 4
6 8 10 12 14 16 18 20
Check It Out! Example 1b
Solve the inequality and graph the solutions.
–50 ≥ 5q
Since q is multiplied by 5, divide
both sides by 5.
–10 ≥ q
–15
–10
–5
0
5
15
Check It Out! Example 1c
Solve the inequality and graph the solutions.
Since g is multiplied by ,
multiply both sides by the
reciprocal of .
g > 36
36
15
20
25
30
35
40
If you multiply or divide both sides of an
inequality by a negative number, the resulting
inequality is not a true statement! You need to
reverse the inequality symbol to make the
statement true.
This means there is another set of properties
of inequality for multiplying or dividing by a
negative number.
Caution!
Do not change the direction of the inequality
symbol just because you see a negative
sign. For example, you do not change the
symbol when solving 4x < –24.
Example 2A: Multiplying or Dividing by a Negative
Number
Solve the inequality and graph the solutions.
–12x > 84
Since x is multiplied by –12, divide
both sides by –12. Change > to <.
x < –7
–7
–14–12 –10 –8 –6 –4 –2 0
2 4
6
Example 2B: Multiplying or Dividing by a Negative
Number
Solve the inequality and graph the solutions.
Since x is divided by –3, multiply
both sides by –3. Change to .
24  x (or x  24)
10 12 14 16 18 20 22 24 26 28 30
Check It Out! Example 2
Solve each inequality and graph the solutions.
a. 10 ≥ –x
–1(10) ≤ –1(–x)
Multiply both sides by –1 to make x
positive. Change  to .
–10 ≤ x
–10 –8 –6 –4 –2 0
2 4
6 8 10
b. 4.25 > –0.25h
Since h is multiplied by –0.25, divide
both sides by –0.25. Change > to <.
–17 < h
–17
–20–16–12–8 –4 0 4
8 12 16 20
Lesson Quiz
Solve each inequality and graph the solutions.
1. 8x < –24 x < –3
2. –5x ≥ 30
x ≤ –6
3.
4.
x≥6
x > 20
5. A soccer coach plans to order more shirts for
her team. Each shirt costs $9.85. She has $77
left in her uniform budget. What are the
possible number of shirts she can buy?
0, 1, 2, 3, 4, 5, 6, or 7 shirts
Inequalities that contain
more than one operation
require more than one
step to solve.
Example 1A: Solving Multi-Step Inequalities
Solve the inequality and graph the solutions.
45 + 2b > 61
45 + 2b > 61
–45
–
45
2b > 16
b>8
0 2 4
Since 45 is added to 2b,
subtract 45 from both
sides to undo the addition.
Since b is multiplied by 2,
divide both sides by 2 to
undo the multiplication.
6 8 10 12 14 16 18 20
Example 1B: Solving Multi-Step Inequalities
Solve the inequality and graph the solutions.
8 – 3y ≥ 29
8 – 3y ≥ 29
–8
–8
–3y ≥ 21
y ≤ –7
Since 8 is added to –3y,
subtract 8 from both sides to
undo the addition.
Since y is multiplied by –3,
divide both sides by –3 to
undo the multiplication.
Change ≥ to ≤.
–7
–10 –8 –6 –4 –2 0
2 4
6 8 10
Check It Out! Example 1a
Solve the inequality and graph the solutions.
–12 ≥ 3x + 6
–12 ≥ 3x + 6
–6
–6
Since 6 is added to 3x,
subtract 6 from both sides to
undo the addition.
–18 ≥ 3x
Since x is multiplied by 3,
divide both sides by 3 to
undo the multiplication.
–6 ≥ x
–10 –8 –6 –4 –2 0
2 4
6 8 10
Check It Out! Example 1b
Solve the inequality and graph the solutions.
Since x is divided by –2, multiply
both sides by –2 to undo the
division. Change > to <.
x + 5 < –6
–5 –
5
x<–
11
–11
–20
–16
–12
Since 5 is added to x, subtract 5
from both sides to undo the
addition.
–8
–4
0
Check It Out! Example 1c
Solve the inequality and graph the solutions.
1 – 2n ≥ 21
–1
–1
–2n ≥
20
n ≤ –10
Since 1 – 2n is divided by 3,
multiply both sides by 3 to
undo the division.
Since 1 is added to −2n,
subtract 1 from both sides to
undo the addition.
Since n is multiplied by −2,
divide both sides by −2 to
undo the multiplication.
Change ≥ to ≤.
–10
–20
–16
–12
–8
–4
0
To solve more complicated inequalities,
you may first need to simplify the
expressions on one or both sides by using
the order of operations, combining like
terms, or using the Distributive Property.
Example 2A: Simplifying Before Solving Inequalities
Solve the inequality and graph the solutions.
2 – (–10) > –4t
12 > –4t
Combine like terms.
Since t is multiplied by –4, divide
both sides by –4 to undo the
multiplication. Change > to <.
–3 < t(or t > –3)
–3
–10 –8 –6 –4 –2 0
2 4
6 8 10
Example 2B: Simplifying Before Solving Inequalities
Solve the inequality and graph the solutions.
–4(2 – x) ≤ 8
−4(2 – x) ≤ 8
−4(2) − 4(−x)
≤ 8 –8 + 4x ≤ 8
+8
+8
4x ≤
16
Distribute –4 on the left side.
Since –8 is added to 4x, add 8 to
both sides.
Since x is multiplied by 4, divide
both sides by 4 to undo the
multiplication.
x≤4
–10 –8 –6 –4 –2 0
2 4
6 8 10
Example 2C: Simplifying Before Solving Inequalities
Solve the inequality and graph the solutions.
Multiply both sides by 6, the LCD
of the fractions.
Distribute 6 on the left side.
4f + 3 > 2
–3 –3
4f
> –1
Since 3 is added to 4f, subtract
3 from both sides to undo the
addition.
Example 2C Continued
4f > –1
Since f is multiplied by 4, divide
both sides by 4 to undo the
multiplication.
0
Check It Out! Example 2a
Solve the inequality and graph the solutions.
2m + 5 > 52
2m + 5 > 25
–5>–5
2m
> 20
m > 10
0 2 4
Simplify 52.
Since 5 is added to 2m, subtract
5 from both sides to undo the
addition.
Since m is multiplied by 2, divide
both sides by 2 to undo the
multiplication.
6 8 10 12 14 16 18 20
Check It Out! Example 2b
Solve the inequality and graph the solutions.
3 + 2(x + 4) > 3
3 + 2(x + 4) > 3
3 + 2x + 8 > 3
2x + 11 > 3
– 11 – 11
2x
> –8
x > –4
–10 –8 –6 –4 –2 0
2 4
Distribute 2 on the left side.
Combine like terms.
Since 11 is added to 2x,
subtract 11 from both sides
to undo the addition.
Since x is multiplied by 2,
divide both sides by 2 to
undo the multiplication.
6 8 10
Check It Out! Example 2c
Solve the inequality and graph the solutions.
Multiply both sides by 8, the
LCD of the fractions.
Distribute 8 on the right side.
5 < 3x – 2
+2
+2
7 < 3x
Since 2 is subtracted from 3x,
add 2 to both sides to undo
the subtraction.
Check It Out! Example 2c Continued
Solve the inequality and graph the solutions.
7 < 3x
Since x is multiplied by 3, divide
both sides by 3 to undo the
multiplication.
0
2
4
6
8
10
Lesson Quiz: Part I
Solve each inequality and graph the solutions.
1. 13 – 2x ≥ 21 x ≤ –4
2. –11 + 2 < 3p p > –3
3. 23 < –2(3 – t) t > 7
4.
Lesson Quiz: Part II
5. A video store has two movie rental plans.
Plan A includes a $25 membership fee plus
$1.25 for each movie rental. Plan B costs
$40 for unlimited movie rentals. For what
number of movie rentals is plan B less
than plan A?
more than 12 movies
Some inequalities have variable terms on both sides of
the inequality symbol. You can solve these inequalities
like you solved equations with variables on both sides.
Use the properties of inequality to “collect” all the variable
terms on one side and all the constant terms on the other
side.
Example 1A: Solving Inequalities with Variables on Both Sides
Solve the inequality and graph the solutions.
y ≤ 4y + 18
y ≤ 4y + 18
–y –y
0 ≤ 3y + 18
–18
– 18
To collect the variable terms on one side,
subtract y from both sides.
Since 18 is added to 3y, subtract 18 from both
sides to undo the addition.
–18 ≤ 3y
Since y is multiplied by 3, divide both sides
by 3 to undo the multiplication.
–6 ≤ y (or y  –6)
–10 –8 –6 –4 –2
0
2
4
6
8 10
Example 1B: Solving Inequalities with Variables on Both Sides
Solve the inequality and graph the solutions.
4m – 3 < 2m + 6
–2m
– 2m
2m – 3 <
+3
2m
<
+6
+3
9
To collect the variable terms on one side,
subtract 2m from both sides.
Since 3 is subtracted from 2m, add 3 to
both sides to undo the subtraction
Since m is multiplied by 2, divide both
sides by 2 to undo the multiplication.
4
5
6
Check It Out! Example 1a
Solve the inequality and graph the solutions.
4x ≥ 7x + 6
4x ≥ 7x + 6
–7x –7x
To collect the variable terms on one side,
subtract 7x from both sides.
–3x ≥ 6
Since x is multiplied by –3, divide both
sides by –3 to undo the multiplication.
Change ≥ to ≤.
x ≤ –2
–10 –8 –6 –4 –2
0
2
4
6
8 10
Check It Out! Example 1b
Solve the inequality and graph the solutions.
5t + 1 < –2t – 6
5t + 1 < –2t – 6
+2t
+2t
To collect the variable terms on one side,
add 2t to both sides.
7t + 1 < –6
Since 1 is added to 7t, subtract 1 from both
sides to undo the addition.
– 1 < –1
7t
< –7
Since t is multiplied by 7, divide both
sides by 7 to undo the
multiplication.
7t < –7
7
7
t < –1
–5 –4 –3 –2 –1
0
1
2
3
4
5
You may need to simplify one or
both sides of an inequality before
solving it. Look for like terms to
combine and places to use the
Distributive Property.
Example 3A: Simplify Each Side Before Solving
Solve the inequality and graph the solutions.
2(k – 3) > 6 + 3k – 3
2(k – 3) > 3 + 3k
Distribute 2 on the left side of the
inequality.
2k + 2(–3) > 3 + 3k
2k – 6 > 3 + 3k
–2k
– 2k
To collect the variable terms,
subtract 2k from both sides.
–6 > 3 + k
–3
–3
–9 > k
Since 3 is added to k, subtract 3 from both
sides to undo the addition.
Example 3A Continued
–9 > k
–12
–9
–6
–3
0
3
Example 3B: Simplify Each Side Before Solving
Solve the inequality and graph the solution.
0.9y ≥ 0.4y – 0.5
0.9y ≥ 0.4y – 0.5
To collect the variable terms, subtract
0.4y from both sides.
–0.4y –0.4y
0.5y ≥
– 0.5
Since y is multiplied by 0.5, divide both
sides by 0.5 to undo the
multiplication.
0.5y ≥ –0.5
0.5
0.5
y ≥ –1
–5 –4 –3 –2 –1
0
1
2
3
4
5
Check It Out! Example 3a
Solve the inequality and graph the solutions.
5(2 – r) ≥ 3(r – 2)
5(2 – r) ≥ 3(r – 2)
Distribute 5 on the left side of the
inequality and distribute 3 on the right
side of the inequality.
5(2) – 5(r) ≥ 3(r) + 3(–2)
10 – 5r ≥ 3r – 6
+6
+6
Since 6 is subtracted from 3r, add 6 to
both sides to undo the subtraction.
16 − 5r ≥ 3r
+ 5r +5r
16
≥ 8r
Since 5r is subtracted from 16 add 5r to
both sides to undo the subtraction.
Check It Out! Example 3a Continued
16 ≥ 8r
Since r is multiplied by 8, divide both
sides by 8 to undo the
multiplication.
2≥r
–6
–4
–2
0
2
4
Check It Out! Example 3b
Solve the inequality and graph the solutions.
0.5x – 0.3 + 1.9x < 0.3x + 6
2.4x – 0.3 < 0.3x + 6
Simplify.
2.4x – 0.3 < 0.3x + 6
Since 0.3 is subtracted from
2.4x, add 0.3 to both sides.
+ 0.3
2.4x
–0.3x
2.1x
+ 0.3
< 0.3x + 6.3
–0.3x
<
Since 0.3x is added to 6.3,
subtract 0.3x from both
sides.
6.3
Since x is multiplied by 2.1,
divide both sides by 2.1.
x<3
Check It Out! Example 3b Continued
x<3
–5 –4 –3 –2 –1
0
1
2
3
4
5
There are special cases of inequalities called identities
and contradictions.
Example 4A: Identities and Contradictions
Solve the inequality.
2x – 7 ≤ 5 + 2x
2x – 7 ≤ 5 + 2x
–2x
–2x
–7 ≤ 5

Subtract 2x from both sides.
True statement.
The inequality 2x − 7 ≤ 5 + 2x is an identity. All values of
x make the inequality true. Therefore, all real numbers
are solutions.
Example 4B: Identities and Contradictions
Solve the inequality.
2(3y – 2) – 4 ≥ 3(2y + 7)
2(3y – 2) – 4 ≥ 3(2y + 7)
Distribute 2 on the left side and 3
on the right side.
2(3y) – 2(2) – 4 ≥ 3(2y) + 3(7)
6y – 4 – 4 ≥ 6y + 21
6y – 8 ≥ 6y + 21
–6y
–6y
–8 ≥ 21
Subtract 6y from both sides.

False statement.
No values of y make the inequality true. There are no
solutions.
Check It Out! Example 4a
Solve the inequality.
4(y – 1) ≥ 4y + 2
4(y – 1) ≥ 4y + 2
Distribute 4 on the left side.
4(y) + 4(–1) ≥ 4y + 2
4y – 4 ≥ 4y + 2
–4y
–4y
–4 ≥ 2
Subtract 4y from both sides.

False statement.
No values of y make the inequality true. There are no
solutions.
Check It Out! Example 4b
Solve the inequality.
x–2<x+1
x–2<x+1
–x
–x
–2 < 1

Subtract x from both sides.
True statement.
All values of x make the inequality true.
All real numbers are solutions.
Lesson Quiz: Part I
Solve each inequality and graph the solutions.
1. t < 5t + 24
t > –6
2. 5x – 9 ≤ 4.1x – 81
3. 4b + 4(1 – b) > b – 9
x ≤ –80
b < 13
Lesson Quiz: Part II
4. Rick bought a photo printer and supplies for $186.90,
which will allow him to print photos for $0.29 each. A
photo store charges $0.55 to print each photo. How
many photos must Rick print before his total cost is
less than getting prints made at the photo store?
Rick must print more than 718 photos.
Lesson Quiz: Part III
Solve each inequality.
5. 2y – 2 ≥ 2(y + 7)
contradiction, no solution
6. 2(–6r – 5) < –3(4r + 2)
identity, all real numbers
The inequalities you have seen so far are
simple inequalities. When two simple
inequalities are combined into one
statement by the words AND or OR, the
result is called a compound inequality.
Example 1: Chemistry Application
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
6.0
is less than or
equal to
≤
pH level
p
is less than or
equal to
≤
6.5
6.0 ≤ p ≤ 6.5
5.9
6.0
6.1
6.2
6.3
6.4
6.5
6.5
Check It Out! Example 1
The free chlorine 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.
Let c be the chlorine level of the pool.
1.0
is less than or
equal to
1.0
≤
chlorine
c
is less than or
equal to
≤
3.0
1.0 ≤ c ≤ 3.0
0
1
2
3
4
5
6
3.0
In this diagram, oval A represents some integer solutions of x <
10 and oval B represents some integer solutions of x > 0. The
overlapping region represents numbers that belong in both
ovals. Those numbers are solutions of both x < 10 and x > 0.
You can graph the solutions of a compound inequality
involving AND by using the idea of an overlapping region.
The overlapping region is called the intersection and shows
the numbers that are solutions of both inequalities.
Example 2A: Solving Compound Inequalities Involving AND
Solve the compound inequality and graph the solutions.
–5 < x + 1 < 2
Since 1 is added to x, subtract 1 from
each part of the inequality.
–5 < x + 1 < 2
–1
–1–1
–6 < x < 1
Graph –6 < x.
Graph x < 1.
Graph the intersection by finding
where the two graphs overlap.
–10 –8 –6 –4 –2
0
2
4
6
8 10
Example 2B: Solving Compound Inequalities Involving AND
Solve the compound inequality and graph the solutions.
8 < 3x – 1 ≤ 11
8 < 3x – 1 ≤ 11
+1
+1 +1
Since 1 is subtracted from 3x, add 1 to
each part of the inequality.
9 < 3x ≤ 12
3<x≤4
Since x is multiplied by 3, divide each part
of the inequality by 3 to undo the
multiplication.
Example 2B Continued
Graph 3 < x.
Graph x ≤ 4.
–5 –4 –3 –2 –1
0
1
2
3
4
5
Graph the intersection by
finding where the two graphs
overlap.
Check It Out! Example 2a
Solve the compound inequality and graph the solutions.
–9 < x – 10 < –5
Since 10 is subtracted from x, add 10 to
each part of the inequality.
–9 < x – 10 < –5
+10
+10 +10
1<x<5
Graph 1 < x.
Graph x < 5.
–5 –4 –3 –2 –1
0
1
2
3
4
5
Graph the intersection by finding
where the two graphs overlap.
Check It Out! Example 2b
Solve the compound inequality and graph the solutions.
–4 ≤ 3n + 5 < 11
–4 ≤ 3n + 5 < 11
–5
–5 –5
–9 ≤ 3n <
Since 5 is added to 3n, subtract 5 from
each part of the inequality.
6
Since n is multiplied by 3, divide each
part of the inequality by 3 to undo
the multiplication.
–3 ≤ n < 2
Graph –3 ≤ n.
Graph n < 2.
–5 –4 –3 –2 –1
0
1
2
3
4
5
Graph the intersection by
finding where the two
graphs overlap.
In this diagram, circle A represents some integer solutions
of x < 0, and circle B represents some integer solutions of x
> 10. The combined shaded regions represent numbers that
are solutions of either x < 0 or x >10.
You can graph the solutions of a compound inequality
involving OR by using the idea of combining regions. The
combine regions are called the union and show the numbers
that are solutions of either inequality.
>
Example 3A: Solving Compound Inequalities Involving OR
Solve the inequality and graph the solutions.
8 + t ≥ 7 OR 8 + t < 2
8 + t ≥ 7 OR 8 + t < 2
–8
–8 –8
−8
t ≥ –1 OR
Solve each simple inequality.
t < –6
Graph t ≥ –1.
Graph t < –6.
–10 –8 –6 –4 –2
0
2
4
6
8 10
Graph the union by combining the
regions.
Example 3B: Solving Compound Inequalities Involving OR
Solve the inequality and graph the solutions.
4x ≤ 20 OR 3x > 21
4x ≤ 20 OR 3x > 21
Solve each simple inequality.
x ≤ 5 OR x > 7
Graph x ≤ 5.
Graph x > 7.
–10 –8 –6 –4 –2
0
2
4
6
8 10
Graph the union by combining the
regions.
Check It Out! Example 3a
Solve the compound inequality and graph the solutions.
2 +r < 12 OR r + 5 > 19
2 +r < 12 OR r + 5 > 19
–2
–2
–5 –5
Solve each simple
inequality.
r < 10 OR r > 14
Graph r < 10.
Graph r > 14.
–4 –2 0
2
4
6
8 10 12 14 16
Graph the union by combining th
regions.
Check It Out! Example 3b
Solve the compound inequality and graph the solutions.
7x ≥ 21 OR 2x < –2
7x ≥ 21 OR 2x < –2
x≥3
OR
Solve each simple
inequality.
x < –1
Graph x ≥ 3.
Graph x < −1.
–5 –4 –3 –2 –1
0
1
2
3
4
5
Graph the union by combining th
regions.
Every solution of a compound inequality involving AND must
be a solution of both parts of the compound inequality. If no
numbers are solutions of both simple inequalities, then the
compound inequality has no solutions.
The solutions of a compound inequality involving OR are not
always two separate sets of numbers. There may be numbers
that are solutions of both parts of the compound inequality.
Example 4A: Writing a Compound Inequality
from a Graph
Write the compound inequality shown by the graph.
The shaded portion of the graph is not between two values, so the compound
inequality involves OR.
On the left, the graph shows an arrow pointing left, so use
either < or ≤. The solid circle at –8 means –8 is a solution so use ≤.
x ≤ –8
On the right, the graph shows an arrow pointing right, so use either > or ≥. The
empty circle at 0 means that 0 is not a solution, so use >.
x>0
The compound inequality is x ≤ –8 OR x > 0.
Example 4B: Writing a Compound Inequality from a Graph
Write the compound inequality shown by the graph.
The shaded portion of the graph is between the values –2 and 5, so the
compound inequality involves AND.
The shaded values are on the right of –2, so use > or ≥. The empty circle at –
2 means –2 is not a solution, so use >.
m > –2
The shaded values are to the left of 5, so use < or ≤. The empty circle at 5
means that 5 is not a solution so use <.
m<5
The compound inequality is m > –2 AND m < 5 (or -2 < m
< 5).
Check It Out! Example 4a
Write the compound inequality shown by the graph.
The shaded portion of the graph is between the values –9 and –2, so the
compound inequality involves AND.
The shaded values are on the right of –9, so use > or . The empty circle at –
9 means –9 is not a solution, so use >.
x > –9
The shaded values are to the left of –2, so use < or ≤. The empty circle at –2
means that –2 is not a solution so use <.
x < –2
The compound inequality is –9 < x AND x < –2
(or –9 < x < –2).
Check It Out! Example 4b
Write the compound inequality shown by the graph.
The shaded portion of the graph is not between two values, so the compound
inequality involves OR.
On the left, the graph shows an arrow pointing left, so use
either < or ≤. The solid circle at –3 means –3 is a solution, so use ≤.
x ≤ –3
On the right, the graph shows an arrow pointing right, so use either > or ≥. The
solid circle at 2 means that 2 is a solution, so use ≥.
x≥2
The compound inequality is x ≤ –3 OR x ≥ 2.
Lesson Quiz: Part I
1. The target heart rate during exercise for a 15 year-old is
between 154 and 174 beats per minute inclusive. Write a
compound inequality to show the heart rates that are
within the target range. Graph the solutions.
154 ≤ h ≤ 174
Lesson Quiz: Part II
Solve each compound inequality and graph the solutions.
2. 2 ≤ 2w + 4 ≤ 12
–1 ≤ w ≤ 4
3. 3 + r > −2 OR 3 + r < −7
r > –5 OR r < –10
Lesson Quiz: Part III
Write the compound inequality shown by each graph.
4.
x < −7 OR x ≥ 0
5.
−2 ≤ a < 4
Now You Try…
Solve and Graph the Compound Inequality
1. -3 < x + 2 < 7
-5 < x < 5
-5
0
5
0
2
4
0
2
4
4
6
8
1
3
5
2. x – 1 < -1 OR x – 5 > -1
x < 0 OR x > 4
3. 2 < x + 2 < 5
0<x<3
4. 11 < 2x + 3 < 21
4<x<9
5. n + 2 < 3 OR n + 3 > 7
x < 1 OR x > 4
Write the compound inequality shown by the graph.
The compound inequality is x ≤ –8 OR x > 0.
Write the compound inequality shown by the graph.
The compound inequality is –9 < x AND x < –2
(or –9 < x < –2).
In Summary:
•
Two inequalities that are combined into one statement by the word AND or OR is
called a compound inequality.
•
If it contains the word AND it is split into two equations and the graph is in
between two points.
•
If it contains the word OR the graphs go in opposite directions from each point.
If you have a  or  you are working with a conjunction, an ‘and’
statement.
Remember: “Less thand”
If you have a  or  you are working with a disjunction, an ‘or’
statement.
Remember: “Greator”
Graphing Inequalities
Parentheses/bracket method :
If the variable is on the
left, the arrow points
the same direction as
the inequality.
Parentheses: endpoint is not included <, >
Bracket: endpoint is included ≤, ≥
x<2
x≥2
Open Circle/closed circle method:
Open Circle: endpoint is not included <, >
Closed Circle: endpoint is included ≤, ≥
x<2
x≥2
Inequalities – Interval Notation
[( smallest, largest )]
Parentheses: endpoint is not included <, >
Bracket: endpoint is included ≤, ≥
Infinity: always uses a parenthesis
x<2
( –∞, 2)
x≥2
[2, ∞)
4<x<9
3-part
inequality
(4, 9)
Inequalities – Set-builder Notation
{variable | condition }
pipe
{ x | x  5}
The set of all x such that x is greater
than or equal to 5.
x<2
( –∞, 2)
{ x |x < 2 }
[2, ∞)
{ x | x ≥ 2}
(4, 9)
{ x | 4 < x < 9}
x≥2
4<x<9
Inequalities
Graph, then write interval notation and setbuilder notation.
x≥5
[
Interval Notation:
[ 5, ∞)
Set-builder Notation: { x | x ≥ 5}
x < –3
)
Interval Notation:
(– ∞, –3)
Set-builder Notation: { x | x < –3 }
Inequalities
Graph, then write interval notation and setbuilder notation.
1<a<6
(
Interval Notation:
)
( 1, 6 )
Set-builder Notation: { a | 1 < a < 6 }
–7 < x ≤ 3
(
Interval Notation:
Set-builder Notation:
]
(– 7, –3]
{ x | –7 < x ≤ 3 }
Solving Inequalities
Solve then graph the solution and write it in
interval notation and set-builder notation.
3x  4  7
4 4
Don’t write = !
3x  3
3
(
3
x 1
Interval Notation:
Set-builder Notation:
( 1, ∞ )
{x|x>1}
Solving Inequalities
Solve then graph the solution and write it in
interval notation and set-builder notation.
4  9k  4k  19
 4k
 4k
4  5k  19
4
4
 5k  15
5 5
]
k  3
Interval Notation:
Set-builder Notation:
(– ∞, –3 ]
{ k | k ≤ –3 }
Solving Inequalities
Solve then graph the solution and write it in
interval notation and set-builder notation.
5
 p  10
3
5 

3
p   3 10
 3 
 5p  30
5
5
)
p6
Interval Notation:
Set-builder Notation:
(– ∞, 6 )
{p|p<6}
Solving Inequalities
Solve then graph the solution and write it in interval notation
and set-builder notation.
1
1
6m  7  3m  1
5
2
1
1


10 6m  7  10 3m  1
 5
2
26m  7  53m  1
12m  14  15m  5
 15m
 12m
 12m
 14  3m  5
5
5
 9  3m
3
3m
3
 15m
 3m  14  5
 14
 3m  9
3
12m  14  15m  5
3
m  3
[
 14
Interval Notation: [– 3, ∞ )
Set-builder Notation: { m | m ≥ – 3 }
Solving Absolute Value
Equations & Inequalities
Absolute Value (of x)
•
•
•
•
Symbol lxl
The distance x is from 0 on the number line.
Always positive
Ex: l-3l=3
-4
2
-3
-2
-1
0
1
Ex: x = 5
• What are the possible values of x?
x=5
or
x = -5
To solve an absolute value equation:
ax+b = c, where c>0
To solve, set up 2 new equations, then solve
each equation.
ax+b = c or ax+b = -c
** make sure the absolute value is by itself
before you split to solve.
Ex: Solve 6x-3 = 15
6x-3 = 15 or 6x-3 = -15
6x = 18 or 6x = -12
x = 3 or x = -2
* Plug in answers to check your solutions!
Ex: Solve 2x + 7 -3 = 8
Get the abs. value part by itself first!
2x+7 = 11
Now split into 2 parts.
2x+7 = 11 or 2x+7 = -11
2x = 4 or 2x = -18
x = 2 or x = -9
Check the solutions.
Solving Absolute Value Inequalities
1. ax+b < c, where c>0
Becomes an “and” problem
Changes to: –c<ax+b<c
2. ax+b > c, where c>0
Becomes an “or” problem
Changes to: ax+b>c or ax+b<-c
Ex: Solve & graph.
4 x  9  21
• Becomes an “and” problem
 21  4x  9  21
12  4x  30
15
3 x 
2
-3
7
8
Solve & graph.
3x  2  3  11
• Get absolute value by itself first.
3x  2  8
• Becomes an “or” problem
3x  2  8 or 3x  2  8
3x  10 or
3x  6
10
x
or x  2
3
-2
3
4
Competition Problems
How many integral values of
x satisfy the inequality:
x – 4 < 5x + 3 < x + 30
Answer:
{-1,0,1,2,3,4,5,6}
8
Solve the inequality:
5(k – 4) – 2(k + 6) ≥ 4(k + 1) – 1
Answer:
-35 ≥ k
Solve for x:
8 – 3x > x + 20
Answer:
-3 > x
Solve:
|3x + 4| – 13 < 2x – 7
Answer:
-2 < x < 2
Solve for x:
–7x +18 ≤ 9 + 5x
Answer:
x ≥ 3/4
Solve the compound inequality:
5x – 7 ≤ 4x – 5 and –3x +15 ≤ 9
Answer:
x=2
Find the sum of the solution(s):
|2x – 1| = 3x + 2
Answer:
-1/5
(Check your answers! The solution
x=-3 does not work!)
Solve:
5 x  4  3x  5
Answer:
1 9 

  ,     ,  
8 2 


Means union (or)