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Section 2.2: Solving Linear Inequalities
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o Solving linear inequalities.
o Solving compound linear inequalities.
o Solving absolute values inequalities.
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HAWKES LEARNING SYSTEMS math courseware specialists
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o If the equality symbol in a linear equation is replaced
linear inequality .
o For example, a linear inequality in is of the form ax b c x o The solution of a linear inequality typically consists of some interval of real numbers described in set notation, graphically or with interval notation.
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Cancellation Properties for Inequalities are true for all inequalities.
Property
A
B
A C B C
If C
0, A
B
A
C
B
C
If C
0, A
B
A
C
B
C
Description
Adding the same quantity to both sides of an inequality results in an equivalent inequality.
If both sides of an inequality are multiplied by a positive quantity , the sign of the inequality is unchanged .
If both sides of an inequality are multiplied by a negative quantity , the sign of the inequality is reversed .
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Solve the following linear inequality.
Step 1: Distribute.
Step 2: Combine like terms.
x
2)
x )
x
x
18
x the reversal of the inequality sign.
5 x
20 x
4
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Solve the following linear inequality.
2 4
7
7
v
14( v
2
v
2
5
5
5)
9
9
4 v v
9 v
4
4
14 v
70
9 v
5 v
70 v
14
Solution is (
,14)
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o The solutions in the previous examples were described using interval notation, but graphing can also be used to describe solutions. o Like in interval notation, parentheses are used when endpoints are not included in the interval and brackets are used when the endpoints are included in the interval.
o For example,
4,
is graphed as follows:
,14
is graphed as follows:
4
14
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A compound inequality is a statement containing two inequality symbols, and can be interpreted as two distinct inequalities joined by the word “and”.
For example, in a course where the grade depends solely on the grades of 5 exams, the following compound inequality could be used to determine the final exam grade needed to score a B in the course.
80
92 65 71 80
x
90
5
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Solve the compound inequality from the previous slide.
80
92 65 71 80
5
x
90
Step 1: Multiply all 400
308 x 450
92 142 from all sides.
Note: If this compound inequality relates to test scores, as
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Solve the compound inequality.
12
x
15
x
15
Note: each inequality is reversed since we are dividing by a negative number!
5 x
25
2 x 5
5
2 x
5
Solution is
5,
2
5
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An absolute value inequality is an inequality in which some variable expression appears inside absolute value symbols. x zero on the real number line. This means that absolute value inequalities can be written without absolute x
a
a and x
a
x
a or x
a
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Example 5: Solving Absolute Value Inequalities
Solve the following absolute value inequality.
5 y 3 2 9
5 y 3 7
5 y
7 Step 2: Rewrite the inequality without absolute values.
Step 3: Solve as compound inequality.
5 y
10
4
2
5
5
4
Solution is , 2
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Example 6: Solving Absolute Value Inequalities
Solve the following absolute value inequality and graph the solution.
x
8
x
8
2 x
14 x
7 or or
6
2 x
8
2 x
2 x
1
1 or
Solution is
, 1
7,
7
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Example 7: Solving Absolute Value Inequalities
Solve the following absolute value inequality
10 x 7 4 x 7 6
Solution is
The solution set is the empty set, as it is impossible for the absolute value of any expression to be negative.
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o Many real-world applications leading to inequalities involve notions such as “is not greater than”, “at least as great as”, “does not exceed”, and so on. o Phrases such as these all have precise mathematical translations that use one of the four inequality
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Sign Phrase
“is greater than”
Example
The average temperature in New
Mexico during the month of May is greater than 75. x
75
“at least as great as” The average price for a large screen digital TV is at least $9000. x
9000
“is less than” The number of skiers on the mountain is less than 100. x
100
“is not greater than” In a quality test, the number of defective products cannot be greater than 5. x
5
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The average daily high temperature in Phoenix, Arizona over the course of three days exceeded 80 degrees. Given that the high on the first day was 78 and the high on the third day was 85, what was the minimum high temperature on the second day?
78 85
80
3
163 240 x
77
The high temperature on the second day was greater than
77.
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As a test for quality at a plant manufacturing silicon wafers, a random sample of 10 batches of 1000 wafers each must not detect more than 5 defective wafers per batch on average. In the first 9 batches tested, the average number of defective wafers per batch is found to be 4.89. What is the maximum number of defective wafers that can be found in the 10 th batch for the plant to pass the quality test?
x
5
10
44.01
50 x
5.99