Math 0031
Section 1.6
Page 1 of 4
1.6 Linear Inequalities
Principles For Solving Inequalities
The Additional Principle
a < b ⇔ a + c < b + c.
The Multiplication Principle
• a < b and c > 0 ⇔ ac < bc.
• a < b and c < 0 ⇔ ac > bc.
Similar statements hold for a > b, a ≤ b, a ≥ b.
Example Solve the inequality 4 − 2x ≤ 5x + 18
Solution:
4 − 2x ≤ 5x + 18
⇔ 4 − 7x ≤ 18 (using the additional principle by adding −5x)
⇔ −7x ≤ 14 (using the additional principle by adding −4)
⇔ x ≥ −2 (using the multiplication principle by multiplying by − 17 ).
The solution set is {x | x ≥ −2} or x ∈ [2, ∞).
The graph (geometric representation) of the solution is
2
−3
−2
−1
0
1
2
3
4
5
6
7
Example Solve the inequality 4x − 1 > 6x + 2
Solution:
4x − 1 > 6x + 2
⇔ 4x > 6x + 3 (using the additional principle by adding 1)
⇔ −2x > 3 (using the additional principle by adding −6x)
⇔ x < − 23
(using the multiplication principle by multiplying by − 12 ).
The solution set is {x | x < − 32 } or x ∈ −∞, − 32 .
The graph (geometric representation) of the solution is
− 32
−6
−5
−4
−3
−2
−1
0
1
2
3
Math 0031
Section 1.6
Page 2 of 4
x−1
Example Find the domain of the function f (x) = √
.
5 − 3x
Solution: Because of the square root we have 5 − 3x ≥ 0 (a radicand under a radical must
be nonnegative). Because of the denominator we have 5 − 3x 6= 0.
Together these two inequalities give the inequality
5 − 3x > 0
⇔ −3x > 5 (using the additional principle by adding −5)
⇔ x < − 35
(using the multiplication principle by multiplying by − 13 ).
The solution set is {x | x < − 32 } or x ∈ −∞, − 53 .
The graph (geometric representation) of the solution is
− 53
−6
−5
−4
−3
−2
−1
0
1
2
3
Compound Inequalities
A compound inequality is made of two inequalities joined by the word and or the word or.
The type of the compound inequality like
a < b and b < c
is called a conjunction. Its solution is represented by an intersection of solution sets of each
inequality.
It can be abbreviated to the double inequality a < b < c.
The type of the compound inequality like
a < b or b < c
is called a disjunction. Its solution is represented by a union of solution sets of each inequality.
It cannot be abbreviated.
Example Solve the double inequality x − 2 ≤ 3x + 2 < 5
Solution: We have to solve two inequalities x − 2 ≤ 3x + 2 and 3x + 2 < 5 and find the
intersection of the corresponding solution sets.
x − 2 ≤ 3x + 2 ⇔ x ≤ 3x + 4
⇔ −2x ≤ 4 ⇔ x ≥ −2.
The solution set is {x | x ≥ 2} or x ∈ [−2, ∞).
Math 0031
Section 1.6
Page 3 of 4
The graph of the solution is
−2
−6
−5
−4
−3
3x + 2 < 5 ⇔ 3x < 3
−2
−1
0
1
2
3
4
2
3
4
⇔ x < 1.
The solution set is {x | x < 1} or x ∈ (−∞, 1).
The graph of the solution is
1
−6
−5
−4
−3
−2
−1
0
1
The intersection of two solution sets is
[−2, ∞) ∩ (−∞, 1) = [−2, 1)
Therefore, the solution set of the double inequality is {x | − 2 ≤ x < 1} or x ∈ [−2, 1).
The graph of the solution is
−2
−6
−5
−4
−3
−2
1
−1
0
1
2
3
4
Example Solve 4x + 3 < −1 or 2 ≤ 5x − 8
Solution: We have to solve two inequalities 4x + 3 < −1 and 2 ≤ 5x − 8 and find the union
of the corresponding solution sets.
4x + 3 < −1 ⇔ 4x < −4
⇔ x < −1.
The solution set is {x | x < −1} or x ∈ (−∞, −1).
The graph of the solution is
−1
−5
−4
−3
−2
−1
2 ≤ 5x − 8 ⇔ 2 − 5x ≤ −8
0
1
2
⇔ −5x ≤ −10
The solution set is {x | x ≥ 2} or x ∈ [2, ∞).
The graph of the solution is
3
4
5
⇔ x ≥ 2.
6
Math 0031
Section 1.6
Page 4 of 4
2
−5
−4
−3
−2
−1
0
1
2
3
4
5
6
The union of two solution sets is {x | x < 1 or x ≥ 2} or x ∈ (−∞, 1) ∪ [2, ∞).
The graph of the solution is
−1
−5
−4
−3
−2
−1
2
0
1
2
3
4
5
6
Applications
Example (Moving Costs) Acme Movers charges $200 plus $45 per hour to move a household
across town. Leo’s Movers charges $70 per hour. For what lengths of time does it cost less to
hire Leo’s Movers?
Solution: Let x be length of time when costs of both companies are the same. At this moment
of time the cost to hire Acme Movers is 200 + 45x and the cost to hire Leo’s Movers is 70x.
If it is less in cost to hire Leo’s Movers then 70x has to be smaller then 200 + 45x.
So, we get the inequality 70x < 200 + 45x.
Then 70x < 200 + 45x ⇔ 25x < 200 ⇔ x < 8.
Assuming that time is never negative we get the solution set {x | 0 ≤ x < 8} or x ∈ [0, 8).
The graph of the solution is
0
0
8
2
4
6
8
10
12
time, in hours