Ultimate guide to linear inequalities

Solving Linear Inequalities

1.4 Sets, Inequalities, and Interval Notation
1.5 Intersections, Unions, and Compound Inequalities

1.6 Absolute-Value Equations and Inequalities

1.4

Sets, Inequalities, and Interval
Notation
OBJECTIVES

a Determine whether a given number is a solution of an
inequality.
b Write interval notation for the solution set or the graph
of an inequality.
c Solve an inequality using the addition principle and the
multiplication principle and then graph the inequality.
d Solve applied problems by translating to inequalities.

1.4

Notation

Inequality
An inequality is a sentence containing

1.4

Notation

Solution of an Inequality
Any replacement or value for the variable that makes
an inequality true is called a solution of the inequality.
The set of all solutions is called the solution set. When
all the solutions of an inequality have been found, we
say that we have solved the inequality.

1.4

Notation

Determine whether a given number is a solution of an
a
inequality.
EXAMPLE 1 Determine whether the given number is a
solution of the inequality.

1.4

Notation

a
inequality.
EXAMPLE 1 Solution

We substitute 5 for x and get 5 + 3 < 6, or 8 < 6, a false
sentence. Therefore, 5 is not a solution.

1.4

Notation

a
inequality.
EXAMPLE 3 Determine whether the given number is a
solution of the inequality.

1.4

Notation

a
inequality.
EXAMPLE 3 Solution

We substitute –3 for x and get
or
a true sentence. Therefore, –3 is a
solution.

1.4
b

Notation

Write interval notation for the solution set or the
graph of an inequality.

The graph of an inequality is a drawing that represents its
solutions.

1.4

Notation

b
EXAMPLE 4 Graph on the number line.

1.4

Notation

b
EXAMPLE 4 Solution

The solutions are all real numbers less than 4, so we
shade all numbers less than 4 on the number line. To
indicate that 4 is not a solution, we use a right
parenthesis “)” at 4.

1.4
b

Notation


1.4

Notation

b
EXAMPLE
Write interval notation for the given set or
graph.

1.4

Notation

b
Solution
EXAMPLE

1.4
c

Notation

Solve an inequality using the addition principle and the

Two inequalities are equivalent if they have the same
solution set.

1.4

Notation

The Addition Principle for Inequalities

1.4

Notation

c
EXAMPLE 10 Solve and graph.

1.4

Notation

c
EXAMPLE 10 Solution

We used the addition principle to show that the
inequalities x + 5 > 1 and x > –4 are equivalent. The
solution set is
and consists of an infinite
number of solutions. We cannot possibly check them all.

1.4

Notation

c
EXAMPLE 10 Solution
Instead, we can perform a partial check by substituting
one member of the solution set (here we use –1) into the
original inequality:

1.4

Notation

c
EXAMPLE 10 Solution

Since 4 > 1 is true, we have a partial check. The solution
set is
or
The graph is as follows:

1.4

Notation

c

1.4

Notation

c
EXAMPLE 11 Solution

1.4

Notation

c
EXAMPLE 11 Solution
The inequalities
and
have the same
meaning and the same solutions.
The solution set is
or more commonly,
Using interval notation,
we write that the solution set is
The graph is as follows:

1.4

Notation

The Multiplication Principle for Inequalities
For any real numbers a and b, and any positive number c:

For any real numbers a and b, and any negative number
c:

Similar statements hold for

1.4

Notation

The multiplication principle tells us that when we
multiply or divide on both sides of an inequality by a
negative number, we must reverse the inequality symbol
to obtain an equivalent inequality.

1.4

Notation

c

1.4

Notation

c
EXAMPLE 13 Solution

1.4

Notation

c
EXAMPLE 15 Solve.

1.4

Notation

c
EXAMPLE 15 Solution

1.4
d

Notation

Solve applied problems by translating to inequalities.

1.4

Notation

Translating “At Least” and “At Most”

1.4

Notation

EXAMPLE 16 Cost of Higher Education.
The equation C = 126t + 1293 can be used to estimate the
average cost of tuition and fees at two-year public
institutions of higher education, where t is the number of
years after 2000. Determine, in terms of an inequality, the
years for which the cost will be more than $3000.

1.4

Notation

EXAMPLE 16 Solution

1. Familiarize. We already have a formula. To become
more familiar with it, we might make a substitution for
t. Suppose we want to know the cost 15 yr after 2000,
or in 2015. We substitute 15 for t:
C = 126(15) + 1293 = $3183.

1.4

Notation

EXAMPLE 16 Solution
We see that in 2015, the cost of tuition and fees at twoyear public institutions will be more than $3000. To find
all the years in which the cost exceeds $3000, we could
make other guesses less than 15, but it is more efficient to
proceed to the next step.

1.4

Notation

EXAMPLE 16 Solution

2. Translate. The cost C is to be more than $3000. Thus we
have C > 3000. We replace C with 126t + 1293 to find
the values of t that are solutions of the inequality:
126t + 1293 > 3000.

1.4

Notation

EXAMPLE 16 Solution
3. Solve. We solve the inequality:

1.4

Notation

EXAMPLE 16 Solution
4. Check. A partial check is to substitute a value for t
greater than 13.55. We did that in the Familiarize step
and found that the cost was more than $3000.
5. State. The average cost of tuition and fees at two-year
public institutions of higher education will be more
than $3000 for years more than 13.55 yr after 2000, so
we have

Ultimate guide to linear inequalities

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