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Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 3.1 - 1
3.1 Linear Inequalities in One Variable
Multiplication Property of Inequality
Multiplication Property of Inequality

Using the Multiplication Property of Inequality
Solve and graph the solution:
Check: Substitute –8 for m in the equation 3m = –24.
The result should be a true statement.
This shows –8
is a boundary
point.

Now we have to test a number on each side of –8 to
verify that numbers greater than or equal to –8 make the
inequality true. We choose –9 and –7.
–16 –14 –12 –10 –8 –6 –4 – 2 0 2 4

Check: Substitute – 5 for k in the equation –7k = 35.
The result should be a true statement.
This shows –5
is a boundary
point.

Now we have to test a number on each side of –5 to
verify that numbers less than or equal to –5 make the
inequality true. We choose –6 and –4.
–16 –14 –12 –10 –8 –6 –4 – 2 0 2 4

Solving a Linear Inequality
Steps used in solving a linear inequality are:
Step 1 Simplify each side separately. Clear
parentheses, fractions, and decimals using the
distributive property as needed, and combine
like terms.
Step 2 Isolate the variable terms on one side. Use
the additive property of inequality to get all
terms with variables on one side of the
inequality and all numbers on the other side.
Step 3 Isolate the variable. Use the multiplication
property of inequality to change the inequality to
the form x < k or x > k.

Step 1
Step 2

Step 3
–10 –9 –8 –7 –6 –5 –4 – 3 –2 –1 0

Solving Linear Inequalities with Three Parts
In some applications, linear inequalities have three parts.
When linear inequalities have three parts, it is important
to write the inequalities so that:
1. The inequality symbols point in the same
direction.
2. Both inequality symbols point toward the lesser
numbers.

Solving a Three-Part Inequality
This statement says that x – 2 is greater than or equal to
3 and less than or equal to 7.
To solve this inequality, we need to isolate the variable
x. To do this, we must add 2 to the expression, x – 2.
To produce an equivalent statement, we must also add
2 to the other two parts of the inequality as well.

3 4 5 6 7 8 9 10 11 12 13

0
–1
–2 1 2

Solving Applied Problems Using Linear Inequalities
In addition to the familiar phrases “less than” and
“greater than”, it is important to accurately interpret
the meaning of the following:
Word Expression Interpretation
a is at least b
a is no less than b
a is at most b
a is no more than b

A rectangle must have an area of at least 15 cm2
and no
more than 60 cm2
. If the width of the rectangle is 3 cm,
what is the range of values for the length?
Step 1 Read the problem.
Step 2 Assign a variable. Let L = the length of the
rectangle.
Step 3 Write an inequality. Area equals width times
length, so area is 3L; and this amount must
be at least 15 and no more than 60.

and no
more than 60 cm2
Step 4 Solve.
Step 5 State the answer. In order for the rectangle to
have an area of at least 15 cm2
and no more
than 60 cm2
when the width is 3 cm, the length
must be at least 5 cm and no more than 20 cm.

and no
more than 60 cm2
Step 6 Check. If the length is 5 cm, the area will be
3 • 5 = 15 cm2
; if the length is 20 cm, the
area will be 3 • 20 = 60 cm2
. Any length
between 5 and 20 cm will produce an area
between 15 and 60 cm2
.

You have just purchased a new cell phone. According to
the terms of your agreement, you pay a flat fee of $6 per
month, plus 4 cents per minute for calls. If you want your
total bill to be no more than $10 for the month, how many
minutes can you use?
Step 1 Read the problem.
Step 2 Assign a variable. Let x = the number of
minutes used during the month.

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