Intro to Inequalities
Writing, Graphing, and Solving
Inequalities
An inequality states that two quantities
either are not equal or may not be
equal. An inequality uses one of the
following symbols:
Symbol
<
>
≤
≥
Meaning
Word Phrases
is less than
Fewer than, below
is greater than
is less than or
equal to
More than, above
is greater than
or equal to
At least, no less than
At most, no more than
Read and write inequalities and
graph them on a number line.
Additional Example 1: Writing Inequalities
Write an inequality for each situation.
A. There are at least 15 people in the
waiting room.
“At least” means greater
than or equal to.
B. The tram attendant will allow no more
than 60 people on the tram.
number of people ≥ 15
number of people ≤ 60
“No more than” means
less than or equal to.
Try This: Example 1
Write an inequality for each situation.
A. There are at most 10 gallons of gas in
the tank.
gallons of gas ≤ 10
“At most” means less
than or equal to.
B. There is at least 10 yards of fabric left.
yards of fabric ≥ 10
“At least” means
greater than or equal to.
1
An inequality that contains a variable is an
algebraic inequality. A value of the variable
that makes the inequality true is a solution of
the inequality.
This open circle shows that 5 is not a solution.
a>5
An inequality may have more than one solution.
Together, all of the solutions are called the
solution set.
If the variable is “greater than or equal to” or “less
than or equal to” a number, that number is indicated
with a closed circle.
You can graph the solutions of an inequality on
a number line. If the variable is “greater than”
or “less than” a number, then that number is
indicated with an open circle.
This closed circle shows that 3 is a solution.
b≤3
Additional Example 2A & 2B: Graphing Simple
Inequalities
Graph each inequality.
A. n < 3
–3
–2
–1
0
1
2
3
B. a ≥ –4
–6
–4
–2
0
2
4
6
Try This: Example 2A & 2B
Graph each inequality.
3 is not a solution, so
draw an open circle at
3. Shade the line to
the left of 3.
–4 is a solution, so
draw a closed circle
at –4. Shade the line
to the right of –4.
A. p ≤ 2
–3
–2
–1
0
1
2
3
B. e > –2
–3
–2
–1
0
1
2
3
2 is a solution, so
draw a closed circle at
2. Shade the line to
the left of 2.
–2 is not a solution, so
draw an open circle
at –2. Shade the line
to the right of –2.
A compound inequality is the result of
combining two inequalities. The words and
and or are used to describe how the two
parts are related.
x > 3 or x < –1
–2 < y and y < 4
x is either greater
than 3 or less than–1.
y is both greater than
–2 and less than 4.
y is between –2 and 4.
Writing Math
The compound inequality –2 < y and y < 4
can be written as –2 < y < 4.
2
Additional Example 3A: Graphing Compound
Inequalities
Additional Example 3B: Graphing Compound
Inequalities
Graph each compound inequality.
Graph each compound inequality
A. m ≤ –2 or m > 1
First graph each inequality separately.
B. –3 < b ≤ 0
–3 < b ≤ 0 can be written as the inequalities
–3 < b and b ≤ 0. Graph each inequality separately.
–3 < b
b≤0
m ≤ –2
m>1
•
–6 –4 –2
0
2 4
–6 –4 –2
6
º
0
2
4
6
Then combine the graphs.
–6 –5 –4 –3 –2 – 1
0
1
2
3
4 5
º
–6 –4 –2
•
0
2
4
6
– 6 –4 – 2
0
2
4
6
Then combine the graphs. Remember that
–3 < b ≤ 0 means that b is between –3 and 0, and
includes 0.
6
The solutions of m ≤ –2 or m > 1 are the combined
solutions of m ≤ –2 or m > 1.
–6 –5 –4 –3 –2 – 1
Try This: Example 3A
0
1
2
3
4 5
6
Try This: Example 3B
Graph each compound inequality.
Graph each compound inequality
A. w < 2 or w ≥ 4
First graph each inequality separately.
B. 5 > g ≥ –3
5 > g ≥ –3 can be written as the inequalities
5 > g and g ≥ –3. Graph each inequality separately.
5>g
g ≥ –3
w<2
–6 –4 –2
0
2 4
W≥4
–6 –4 –2
6
0
2
4
6
Then combine the graphs.
–6 –5 –4 –3 –2 – 1
0
1
2
3
4 5
6
The solutions of w < 2 or w ≥ 4 are the combined
solutions of w < 2 or w ≥ 4.
–6 –4 –2
0
2
4
º
•
6
– 6 –4 – 2
0
2
4
6
Then combine the graphs. Remember that
5 > g ≥ –3 means that g is between 5 and –3, and
includes –3.
–6 –5 –4 –3 –2 – 1
0
1
2
3
4 5
6
Solve one-step inequalities by adding or
subtracting.
3
When you add or subtract the same
number on both sides of an inequality,
the resulting statement will still be
true.
–2 < 5
+7 +7
5 < 12
Additional Example 1A: Solving Inequalities by
Adding
Solve. Then graph the solution set on a
number line.
A. n – 7 ≤ 15
n – 7 ≤ 15
+7 +7
n
≤ 22
You can find solution sets of
inequalities the same way you find
solutions of equations, by isolating the
variable.
Add 7 to both sides.
Draw a closed circle at 22 then
shade the line to the left of 22.
–14
–7
Additional Example 1B: Solving Inequalities by
Adding
7
0
14
21
28
Try This: Example 1A
Solve. Then graph the solution set on a
number line.
Solve. Then graph the solution set on a
number line.
B. a – 10 ≥ –3
A. d – 12 ≤ –18
d – 12 ≤ –18
+ 12 + 12
d
≤
–6
a – 10 ≥ –3
+ 10 +10
a
–4
≥
–2
Add 10 to both sides.
7
0
Draw a closed circle at 7.
Then shade the line to the right.
2
4
6
8
10
35
Add 12 to both sides.
Draw a closed circle at –6 then
shade the line to the left of –6.
–8
–6
–4
–2
0
2
4
6
Try This: Example 1B
Solve. Then graph the solution set on a
number line.
B. b – 14 ≥ –8
b – 14 ≥ –8
+ 14 +14
b
–4
≥
–2
Add 14 to both sides.
6
0
Draw a closed circle at 6.
Then shade the line to the right.
2
4
6
8
10
4
To Check Your Answer: You can
see if the solution to an inequality
is true by choosing any number in
the solution set and substituting it
into the original inequality.
Additional Example 2A: Solving Inequalities by
Subtracting
Solve. Check each answer.
A. d + 11 > 6
d + 11 > 6
–11 –11
d
Subtract 11 from both sides.
> –5
Check
d + 11 > 6
Additional Example 2B: Solving Inequalities by
Subtracting
Solve. Check your answer.
B. b + 12 ≤ 19
b + 12 ≤
19
–12
–12
b
≤
Subtract 12 from both sides.
7
?
≤
18
?
≤
?
>
6
11
?
>
6
0 is greater than –5.
Substitute 0 for d.
Try This: Example 2A
Solve. Check each answer.
A. c + 15 > 9
c + 15 > 9
–15 –15
c
Check
b + 12 ≤ 19
6 + 12
0 + 11
Subtract 15 from both sides.
> –6
Check
c + 15 > 9
19
6 is less than 7.
Substitute 6 for b.
19
0 + 15
?
>
9
15
?
>
9
0 is greater than –6.
Substitute 0 for c.
Try This: Example 2B
Solve. Check your answer.
B. a + 15 ≤ 20
a + 15 ≤
20
–15
–15
a
≤
Helpful Hint
When checking your solution, choose a number
in the solution set that is easy to work with.
Subtract 15 from both sides.
5
Check
a + 15 ≤ 20
4 + 15
?
≤
20
19
?
≤
20
4 is less than 5.
Substitute 4 for a.
5
When you multiply or divide both sides
of an inequality by the same positive
number, the statement will still be true.
Solve one-step inequalities by multiplying
or dividing.
–4 < 2
(3)(–4)
<
(3)(2)
–12 < 6
–4 < 2
(–3)(–4)
> (–3)(2)
12 > –6
However, when you multiply or divide
both sides by the same negative
number, you need to reverse the
direction of the inequality symbol
for the statement to be true.
true
Additional Example 1A: Solving Inequalities by
Multiplying
Solve.
A. c ≤ –4
4
c ≤ –4
4
(4)c ≤ (4) (–4)
4
c ≤ –16
Additional Example 1B: Solving Inequalities by
Multiplying
Solve.
B.
t
> 0.3
–4
t > 0.3
–4
t
(–4) < (–4)0.3 Multiply both sides by –4
–4
and reverse the inequality
t < –1.2
symbol.
Multiply both sides by 4.
Try This: Example 1A
Solve.
A. n ≤ –5
6
n ≤ –5
6
(6)n ≤ (6) (–5)
6
Multiply both sides by 6.
n ≤ –30
6
Try This: Example 1B
Solve.
B.
r
> 0.9
–3
r > 0.9
–3
r
(–3) < (–3)0.9 Multiply both sides by –3
–3
and reverse the inequality
r < –2.7
symbol.
Additional Example 2A: Solving Inequalities by
Dividing
Solve. Check your answer.
A. 5a ≥ 23
5a ≥ 23
Divide both sides by 5.
5
5
23
3
a ≥ 5 , or 4
5
Check
5a ≥ 23
?
5(5) ≥ 23
?
25 ≥ 23
Additional Example 2B: Solving Inequalities by
Dividing
Solve. Check your answer.
B. –24b < 192
–24b > 192
–24
–24
b > –8
Divide both sides by –24,
and reverse the inequality
symbol.
Check
–24(0) < 192
?
0 < 192
Try This: Example 2A
Solve. Check your answer.
A. 6b ≥ 25
6b ≥ 25
Divide both sides by 6.
6
6
25
1
b ≥ 6 , or 4
6
Check
–24b < 192
?
5 is greater than 4 3 .
5
Substitute 5 for a.
6b ≥ 25
0 is greater than –8.
Substitute 0 for b.
?
6(6) ≥ 25
?
36 ≥ 25
6 is greater than 4 1 .
6
Substitute 6 for b.
Try This: Example 2B
Solve. Check your answer.
B. –17b < 85
–17b > 85
–17 –17
b > –5
Divide both sides by –17, and
reverse the inequality symbol.
Solve two-step inequalities by multiplying
or dividing.
Check
–17b < 85
?
–17(0) < 85
?
0 < 85
0 is greater than –5.
Substitute 0 for b.
7
When you solve two-step equations,
you can use the order of operations in
reverse to isolate the variable. You can
use the same process when solving
two-step inequalities.
Additional Example 1A: Solving Two-Step
Inequalities
Solve. Then graph the solution set on a
number line.
A. y – 6 > 1
2
y
2 –6>1
+6 +6
Add 6 to both sides.
y
>7
2
y
Multiply both sides by 2.
(2) 2 > (2)7
y > 14
– 21
– 14
–7
0
7
14
21
Additional Example 1C: Solving Two-Step
Inequalities
Solve. Then graph the solution on a number
line.
C. 4y – 5 < 11
4y – 5 < 11
+5 +5
4y
< 16
4y
< 16
4
4
Add 5 to both sides.
Divide both sides by 4.
y<4
–6 –4 –2 0
º
2 4
6
Remember!
Draw a closed circle when the inequality includes
the point and an open circle when it does not
include the point.
Additional Example 1B: Solving Two-Step
Inequalities
Solve. Then graph the solution set on a
number line.
B. m + 8 ≤ 5
–3
m +8≤5
–3
– 8 –8
Subtract 8 from both sides.
m
≤ –3
–3
(–3) m ≥ (–3)(–3) Multiply both sides by –3, and
–3
reverse the inequality symbol.
m≥9
•
–3
0
3
6
9
12
15
Additional Example 1D: Solving Two-Step
Inequalities
Solve. Then graph the solution set on a
number line.
D. –3x + 5 ≤ –4
–3x + 5 ≤ –4
– 5 –5
–3x
≤ –9
–3x ≥ –9
–3
–3
x ≥ 3
–9 –6 –3 0
3 6
Subtract 5 from both sides.
Divide both sides by –3, and
reverse the inequality
symbol.
9
8
Try This: Example 1A
Try This: Example 1B
Solve. Then graph the solution set on a
number line.
A. h + 1 > –1
7
h
7 + 1 > –1
–1
–1
Subtract 1 from both sides.
h
> –2
7
h
(7) 7 > (7)(–2) Multiply both sides by 7.
h > –14
– 21
– 14
–7
0
7
14
21
Try This: Example 1C
Solve. Then graph the solution on a number
line.
C. 2y – 4 > –12
2y – 4 > –12
+4 +4
2y
> –8
2y
> –8
2
2
–6 –4 –2 0
2 4
B. m + 1 ≥ 7
–2
m +1≥7
–2
– 1 –1
Subtract 1 from both sides.
m
≥ 6
–2
(–2) m ≤ (–2)(6) Multiply both sides by –2, and
–2
reverse the inequality symbol.
m ≤ –12
•
–18 –12 –6
0
6
12
18
Try This: Example 1D
Solve. Then graph the solution set on a
number line.
D. –9x + 4 ≤ 31
Add 4 to both sides.
Divide both sides by 2.
y > –4
º
Solve. Then graph the solution set on a
number line.
6
–9x + 4 ≤ 31
– 4 –4
–9x
≤ 27
–9x ≥ 27
–9
–9
x ≥ –3
–9 –6 –3 0
3 6
Subtract 4 from both sides.
Divide both sides by –9,
and reverse the inequality
symbol.
9
9