Slide 2- 1Copyright © 2012 Pearson Education, Inc.Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Copyright © 2012 Pearson Education, Inc.

2.1 Linear Functions and Models

2.2 Equations of Lines

2.3 Linear Equations

2.4 Linear Inequalities

2.5 Absolute value Equations and Inequalities

Linear Functions and Equations

2

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Linear Inequalities

♦ Understand basic terminology related to inequalities

♦ Solve linear inequalities symbolically♦ Solve linear inequalities graphically and

numerically♦ Solve compound inequalities

2.4

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Terminology Related to Inequalities

• Inequalities result whenever the equals sign in an equation is replaced with any one of the symbols ≤, ≥, <, >.

• Examples of inequalities include:• x – 5 > 2x + 3• x2 ≤ 1 – 2x• xy – x < x2 • 5 > 1

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Linear Inequality in One Variable

• A linear inequality in one variable is an inequality that can be written in the form ax + b > 0 where a ≠ 0. (The symbol may be replaced by ≤, ≥, <, > )

• Examples of linear inequalities in one variable:• 5x + 4 ≤ 2 + 3x simplifies to 2x + 2 ≤ 0 1(x – 3) + 4(2x + 1) > 5 simplifies to 7x + 2 > 0

• Examples of inequalities in one variable which are not linear:

• x2 < 1 01

1

x

x

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Interval Notation• The solution to a linear inequality in one variable is typically an

interval on the real number line. See examples of interval notation below.

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What Happens When Both Sides of An Inequality Are Multiplied By A Negative Number?

• Note that 3 < 5, but if both sides are multiplied by 1, in order to produce a true statement the > symbol must be used.

3 < 5

but

3 > 5So when both sides of an inequality are multiplied (or divided) by a negative number the direction of the inequality must be reversed.

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Solving Linear Inequalities Symbolically• The procedure for solving a linear inequality symbolically is the same as

the procedure for solving a linear equation, except when both sides of an inequality are multiplied (or divided) by a negative number the direction of the inequality is reversed.

Example of Solving a Example of Solving aLinear Equation Symbolically Linear Inequality SymbolicallySolve 2x + 1 = x 2 Solve 2x + 1 < x 2 2x x = 2 1 2x x < 2 1 3x = 3 3x < 3 x = 1 xx >> 1 1

Note that we divided both sides by 3 so the

direction was reversed.

In interval notation the solution set is (1,∞).

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Example Solve graphically.

Solution

3 17 2 4x x

[[ 2, 15, 1] by [2, 15, 1] by [ 2, 15, 1]2, 15, 1]

Note that the graphs intersect at the point (8.20, 7.59). The graph of y1 is above the graph of y2 to the right of the point of intersection or when x > 8.20. Thus in interval notation the solution set is (8.20, ∞).

STEP1

STEP2

STEP3

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Comments on Previous Problem

Solve 3 17 2 4.x x

Note that the inequality above becomes y1 > y2 since in Step 1 we let y1 equal the left-hand side of the inequality and y2 equal the right hand side.

To write the solution set of the inequality we are looking for the values of x for which the graph of y1 is above the graph of y2. These are values of x to the right of 8.20, that is values larger than 8.20. Thus the solution is x > 8.20 which is (8.20, ∞) in interval notation.

STEP1

STEP3

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ExampleSolve

Solution

2 7.x πx

[[10, 10, 1] by [10, 10, 1] by [10, 10, 1]10, 10, 1]

Note that the graphs intersect at the point (1.36, 2.72). The graph of y1 is above the graph of y2 to the left of the point of intersection or when x < 1.36. When x ≤ 1.36 the graph of y1 is on or above the graph of y2. Thus in interval notation the solution set is ( ∞, 1.36].

STEP1

STEP2

STEP3

Slide 2- 12Copyright © 2012 Pearson Education, Inc.

Comments on Previous Problem

SolveSolve

Note that the inequality above becomes y1 ≥ y2 since in Step 1 we let y1 equal the left-hand side of the inequality and y2 equal the right hand side.

To write the solution set of the inequality we are looking for the values of x for which the graph of y1 is on or above the graph of y2. These values of x are values less than or equal to 1.36 which is ( ∞, 1.36] in interval notation.

STEP1

STEP3

2 7.x πx

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Example

Solve numerically.

Solution Note that the inequality above becomes y1 ≥ y2 since in Step 1 we let y1 equal the left-

hand side and y2 equal the right hand side.

To write the solution set of the inequality we are looking for the values of x in the table for which y1 is the same or larger than y2. Note

that when x = 1.3, y1 is less than y2; but when

x = 1.4, y1 is larger than y2. By the

Intermediate Value Property, there is a value of x between 1.4 and 1.3 such that y1 = y2.

In order to find an approximation of this value, make a new table in which x is incremented by 0.01 (x is incremented by 0.1 in the table to the left.)

STEP1

STEP2

2 7x πx

Slide 2- 14Copyright © 2012 Pearson Education, Inc.

To write the solution set of the inequality we are looking for the values of x in the table for which y1

is the same as or larger than y2.

Note that when x is approximately 1.36, y1 equals y2 and when x is

smaller than 1.36 y1 is larger than

y2 so the solutions can be written x

≤ 1.36 or (∞, 1.36] in interval notation.

72 πxx

Example continued

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ExampleSuppose the Fahrenheit temperature x miles above the ground level is given by T(x) = 88 – 32x. Determine the altitudes where the air temp is from 300 to 400.

Solution

We must solve the inequality 30 < 88 – 32 x < 40. 8125.15.1

5.18125.12

3

16

2932

48

32

58

483258

8840328830

40328830

x

x

x

x

x

x

x

•The goal is to isolate the variable x in the middle of the three-part inequality

Direction reversed –Divided each side of an inequality by a negative.

Between 1.5 and 1.8215 miles above ground level, the air temperature is between 30 and 40 degrees Fahrenheit.