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Math in Our World
Section 6.5
Solving Linear Inequalities
Learning Objectives
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Graph solution sets for simple inequalities.
Solve linear inequalities in one variable.
Solve three-part linear inequalities.
Solve real-world problems using inequalities.
Linear Inequalities
A linear inequality in one variable is a
statement that can be written in any of the
following four forms:
Ax + B < 0, Ax + B > 0, Ax + B ≤ 0, or Ax + B ≥ 0.
In each case, B can be any real number, and A
can be any real number except zero.
To solve a linear inequality means to find the set of all
numbers that make the inequality a true statement when
substituted in for the variable. That set is called the
solution set for the inequality.
Graphing Inequalities
For an inequality of the form x ≥ 5, every real number
that’s 5 or larger.
To show that the number 5 is included in the solution set, a
solid or closed circle (●) is used.
On the other hand, the solution set for the inequality x > 5
includes all real numbers greater than 5; however, the
actual number 5 is not in the solution set.
To show that the number 5 is not included in the solution
set, an open circle (○) is used.
Graphing Inequalities
For an inequality of the form x ≤ 5, every real number
that’s 5 or smaller.
Again a closed circle indicates that 5 is included.
When taking the intersection of two inequalities, such as
x ≥ 2 and x ≤ 4, it’s more efficient to write it this way:
2 ≤ x ≤ 4.
This is the way we’ll represent an interval between two
numbers.
Graphing Inequalities
EXAMPLE 1
Graphing Solution Sets for
Simple Inequalities
Graph the solution set for each inequality.
(a) x ≤ 10 (b) y > -4 (c) -30 < x ≤ 50
SOLUTION
Solving Linear Inequalities
Addition and Subtraction Properties for Inequalities
You can add or subtract the same real number or
algebraic expression to both sides of an inequality
without changing the solution set. In symbols, if a < b,
then a + c < b + c and a – c < b – c, likewise for >, ≤, ≥.
Solving Linear Inequalities
Multiplication and Division Properties for Inequalities
You can multiply or divide both sides of an inequality by
the same positive real number without changing the
solution set. In symbols, if a < b and c > 0 then
a bc
b and likewise for >, ≤, ≥.
ac < bc and ac

c c
If you multiply or divide both sides of an inequality by the
same negative real number, the direction of the
inequality symbol is reversed. If a < b and c < 0 then
a b
ac > bc and ac bc, and likewise for >, ≤, ≥.
c c
EXAMPLE 2
Solving a Linear Inequality
Solve and graph the solution set for 5x – 9 ≥ 21.
SOLUTION
The solution set is {x | x ≥ 6}. The graph of the solution set is
EXAMPLE 3
Solving a Linear Inequality
Solve and graph the solution set for 16 – 3x > 40.
SOLUTION
The solution set is {x | x < – 8}. The graph is
EXAMPLE 4
Solving a Linear Inequality
Solve and graph the solution set for
4(x + 3) < 2x – 26.
EXAMPLE 4
Solving a Linear Inequality
SOLUTION
The solution set is {x | x < – 19}. The graph is
EXAMPLE 5
Solving a Three-Part
Linear Inequality
Solve and graph the solution set for – 4 < 3 – 2y ≤ 9.
SOLUTION
7

The solution set is 
{x
|
-3
<
x
<
–
8}.
y

3
y



2


Common Phrases in Inequalities
EXAMPLE 6
Applying Inequalities to
Vacation Planning
With the stress of finals behind you, you decide to
plan a vacation to relax a little bit. After poking
around on the Internet, you find a room in the area
you want to visit for $65 per night. Some quick
estimating leads you to conclude that you’ll need
at least $250 for gas, food, beverages, and
entertainment expenses. Upon checking your bank
balance, you decide that you can afford to spend
at most $600 on the trip. How many nights can you
stay?
EXAMPLE 6
Applying Inequalities to
Vacation Planning
SOLUTION
Step 1 Relevant information: Lodging is $65 per night, other
expenses are $250, maximum you can spend is $600. We’re
asked to find the number of nights.
Step 2 Use variable n to represent the number of nights.
Step 3 Translate the relevant information into an inequality.
$65 x number of nights + other expenses is no more than $600
65 x
n
+
250
≤
600
EXAMPLE 6
Applying Inequalities to
Vacation Planning
SOLUTION
Step 4 Solve the inequality.
Step 5 Answer the question. Staying 5.4 nights doesn’t
make sense, so you could stay at most 5 nights.
Step 6 Check: Will 5 nights work?
EXAMPLE 7
Applying Inequalities to the
Cost of Buying Food
Mike is planning to buy lunch for himself and some
coworkers. He decides to buy cheeseburgers and
fries from the value menu—the burgers are $1
each, and the fries cost $0.80. He also needs to
pay 5% of the total in sales tax. What is the largest
number of items he can buy if he wants to buy the
same number of burgers as fries, and he only has
$10 to spend?
EXAMPLE 7
Applying Inequalities to the
Cost of Buying Food
SOLUTION
Step 1 Relevant information: Burgers are $1 each, fries are
$0.80 each; 5% sales tax; same number of burgers and fries;
maximum cost $10.
Step 2 Use variable x to represent both the number of burgers
and the number of fries because those numbers are equal.
Step 3 $ 1 • x is the total cost of burgers, and $0.80x is the total
cost of fries. The tax is 5% of their sum, which is 0.05(x + .80x)
dollars. So the total cost x + 0.80x + 0.05(x + 0.80x), which
must be less than or equal to 10. The inequality is
EXAMPLE 7
Applying Inequalities to the
Cost of Buying Food
SOLUTION
Step 4 Solve the inequality.
Step 5 Mike can’t buy 5.29 burgers, so we round down to 5.
He can buy at most 5 burgers and 5 fries.