Copyright © 2013, 2009, 2006 Pearson Education, Inc. 1 1 Section 2.7 Solving Linear Inequalities...

Section 2.7

Solving LinearInequalities

Objective #1 Graph the solutions of an inequality

on a number line.

Any inequality in the form is called a linear inequality in one variable.

The inequality symbol may be

< less than

> greater than

≤ less than or equal to

≥ greater than or equal to

Linear Inequalities in One Variable

Graphs of Inequalities

Linear Inequalities

EXAMPLE

SOLUTION

Graph the solution of each inequality on a number line and then express the solutions in set-builder notation and interval notation. 72 (b)4 (a) xx

Linear Inequalities

EXAMPLE

SOLUTION

Graph the solution of each inequality on a number line and then express the solutions in set-builder notation and interval notation. 72 (b)4 (a) xx

Linear Inequalities

CONTINUED

Linear Inequalities

CONTINUED

Objective #1: Examples

1a. Graph the solution of the inequality: 4x

The solutions of 4x are all real numbers that are less than 4. They are graphed on a number line by shading all points to the left of 4. The parenthesis at 4 indicates that 4 is not a solution. The arrow shows that the graph extends indefinitely to the left.

1a. Graph the solution of the inequality: 4x

The solutions of 4x are all real numbers that are less than 4. They are graphed on a number line by shading all points to the left of 4. The parenthesis at 4 indicates that 4 is not a solution. The arrow shows that the graph extends indefinitely to the left.

1b. Graph the solution of the inequality: 4 1x

The solutions of 4 1x are all real numbers between –4 and 1, not including 1 but including –4. The square bracket at –4 shows that –4 is a solution. The parenthesis at 1 indicates that 1 is not a solution. Shading indicates the other solutions.

1b. Graph the solution of the inequality: 4 1x

The solutions of 4 1x are all real numbers between –4 and 1, not including 1 but including –4. The square bracket at –4 shows that –4 is a solution. The parenthesis at 1 indicates that 1 is not a solution. Shading indicates the other solutions.

Objective #2 Use interval notation.

Interval Notation

2a. Express the solution set of the inequality in interval notation and graph the interval: 0x The solutions of 0x are all real numbers greater than or equal to 0. Use a square bracket at 0 because 0 is a solution. Since the solution extends indefinitely to the right, use a parenthesis at . 0, Graph:

2b. Express the solution set of the inequality in interval notation and graph the interval: 5x The solutions of 5x are all real numbers less than 5. Since the solution extends indefinitely to the left, use a parenthesis at . Use a parenthesis at 5 because 5 is not a solution. ,5 Graph:

Objective #3 Understand properties used to solve

linear inequalities.

Inequalities

Properties of InequalitiesProperty The Property in Words Example

The Addition Property of Inequality

If a < b, then a + c < b + c

If a < b, then a – c < b - c

If the same quantity is added to or subtracted from both sides of an inequality, the resulting inequality is equivalent to the original one.

2x + 3 < 7

Subtract 3:

2x + 3 – 3 < 7 – 3

Simplify:

2x < 4

Inequalities

Properties of InequalitiesProperty The Property in Words Example

The Positive Multiplication Property of Inequality

If a < b and c is positive, then ac < bc

If a < b and c is positive, then

If we multiply or divide both sides of an inequality by the same positive quantity, the resulting inequality is equivalent to the original one.

2x < 4

Divide by 2:

Simplify:

CONTINUED

Inequalities

CONTINUED

Remember that when you multiply both sides of an inequality by a negative, you must reverse the direction of the inequality symbol.

3. True or False: When we add (or subtract) a negative number to (or from) both sides of an inequality, the direction of the inequality symbol is reversed.

False; This rule applies to multiplication and division.

3. True or False: When we add (or subtract) a negative number to (or from) both sides of an inequality, the direction of the inequality symbol is reversed.

False; This rule applies to multiplication and division.

Objective #4 Solve linear inequalities.

Inequalities

Solving a Linear Inequality1) Simplify the algebraic expression on each side.

2) Use the addition property of inequality to collect all the variable terms on one side and all the constant terms on the other side.

3) Use the multiplication property of inequality to isolate the variable and solve. Reverse the sense of the inequality when multiplying or dividing both sides by a negative number.

4) Express the solution set in set-builder or interval notation and graph the solution set on a number line.

EXAMPLE: Solve 2(x + 3) ≤ 4x + 12

Linear Inequalities

2( 3) 4 12

2 6 4 12 Distribute

2 6 4 4 12 4 Subtract 4 from both sides

2 6 12 Subtract 6 from both sides

x x x x x

6 Divide both sides by 2

EXAMPLE: Solve 2(x + 3) ≤ 4x + 12

Linear Inequalities

2( 3) 4 12

2 6 4 12 Distribute

2 6 4 4 12 4 Subtract 4 from both sides

2 6 12 Subtract 6 from both sides

x x x x x

6 Divide both sides by 2

Linear Inequalities

EXAMPLE

SOLUTION

Solve the linear inequality. Then graph the solution set on a number line.

1335 xx

13515 xx Distribute

1815 xx816

Add 5x to both sidesAdd 1 to both sides

1) Simplify each side.

2) Collect variable terms on one side and constant terms on the other side.

Linear Inequalities

3) Isolate the variable and solve.

-1 0 1 2 3 4 5 6 [

x2 Divide both sides by 8

CONTINUED

4) Express the solution set in set-builder or interval notation and graph the set on a number line.

{x|x 2} = [2, )

Linear Inequalities

EXAMPLE

SOLUTION

Solve the linear inequality. Then graph the solution set on a number line.

Distribute

110110

First we need to eliminate the denominators.

Multiply by LCD = 10

Linear Inequalities

2104 x Add x to both sides

Subtract 10 from both sides

1) Simplify each side. Because each side is already simplified, we can skip this step.

2) Collect variable terms on one side and constant terms on the other side.

110110

xx 2103

CONTINUED 1 2 1

xx 2103

Linear Inequalities

CONTINUED

3) Isolate the variable and solve.

-6 -5 -4 -3 -2 -1 0 1 2 [

2x Divide both sides by 4

4) Express the solution set in set-builder or interval notation and graph the set on a number line.

{x|x -2} = [-2, )

4a. Solve: 6 9x

6 6 9 6

The solution set is ,3 or 3 .x x

4a. Solve: 6 9x

6 6 9 6

4b. Solve: 1

4 4 24

4b. Solve: 1

4 4 24

4c. Solve: 6 18x

6 186 6

The solution set is 3, or 3 .x x

4c. Solve: 6 18x

6 186 6

The solution set is 3, or 3 .x x

Objective #5 Identify inequalities with no solution or true

for all real numbers.

No Solution or Infinitely Many Solutions

Some inequalities have either no solution or have infinitely many solutions. When you attempt to solve such inequalities, all the variables are eliminated and you either get a sentence that is true or false.

If the variables are all eliminated at once and you are left with a true sentence, the original inequality is an identity and is true for all x.

If the variables are all eliminated at once and you are left with a false sentence, the original inequality has no solution.

Linear Inequalities

EXAMPLE

SOLUTION

Solve the linear inequality.

Distribute

Since the result is a false statement, there is no solution.

5x < 5(x – 3)

5x < 5x – 15

Subtract 5x from both sides0 < – 15

5a. Solve: 4( 2) 4 15x x

4( 2) 4 15x x 4 8 4 15

4 4 8 4 4 15

8 15, false

x x x x

There is no solution or .

5a. Solve: 4( 2) 4 15x x

4( 2) 4 15x x 4 8 4 15

4 4 8 4 4 15

8 15, false

x x x x

There is no solution or .

5b. Solve: 3( 1) 2 1x x x

3( 1) 2 1x x x 3 3 3 1

3 3 3 3 3 1

3 1, true

x x x x

The solution is , or is a real number .x x

5b. Solve: 3( 1) 2 1x x x

3( 1) 2 1x x x 3 3 3 1

3 3 3 3 3 1

3 1, true

x x x x

The solution is , or is a real number .x x

Objective #6 Solve problems using linear inequalities.

Linear Inequalities

EXAMPLE

SOLUTIONSOLUTION

You are choosing between two long-distance telephone plans. Plan A has a monthly fee of $15 with a charge of $0.08 per minute for all long-distance calls. Plan B has a monthly fee of $3 with a charge of $0.12 per minute for all long-distance calls. How many minutes of long-distance calls in a month make plan A the better deal?

1) Let x represent one of the quantities. We are looking for the number of minutes of long-distance calls that make Plan A the better deal. Thus,

Let x = the number of minutes of long-distance phone calls.

Linear Inequalities

2) Represent other quantities in terms of x. We are not asked to find another quantity, so we can skip this step.

3) Write an inequality in x that describes the conditions. Plan A is a better deal than Plan B if the monthly cost of Plan A is less than the monthly cost of Plan B. The inequality that represents this is the following, with the information for Plan A on the left side and the information for Plan B on the right side of the inequality.

CONTINUED

15 + 0.08x < 3 + 0.12x

Plan A is less than Plan B.

Linear Inequalities

4) Solve the inequality and answer the question.

Subtract 3 from both sides

CONTINUED

15 + 0.08x < 3 + 0.12x

12 + 0.08x < 0.12x

Subtract 0.08x from both sides15 < 0.04x

Divide both sides by 0.04375 < x

Therefore, Plan A costs less than Plan B when x > 375 long-distance minutes per month.

Linear Inequalities

5) Check the proposed solution in the original wording of the problem. One way to do this is to take a long-distance number of minutes greater than 375 per month to see if Plan A is the better deal. Suppose that the number of long-distance minutes we use is 450 in a month.

CONTINUED

Cost for Plan A = 15 + (450)(0.08) = $51

Plan A has a lower monthly cost, making Plan A the better deal.

Cost for Plan B = 3 + (450)(0.12) = $57

Linear Inequalities

In summary…

Solving a linear inequality is similar to solving a linear equation exceptfor two important differences:

When you solve an inequality, your solution is usually an interval, not a point.

When you multiply both sides of an inequality by a negative number, you must remember to reverse the direction of the inequality sign.

6. To earn a B in a course, you must have a final average of at least 80%. On the first three examinations, you have grades of 82%, 74%, and 78%. If the final examination counts as two grades, what must you get on the final to earn a B in the course?

Let x your grade on the final examination. The average of examinations must be greater than or equal to 80.

6. To earn a B in a course, you must have a final average of at least 80%. On the first three examinations, you have grades of 82%, 74%, and 78%. If the final examination counts as two grades, what must you get on the final to earn a B in the course?

Let x your grade on the final examination. The average of examinations must be greater than or equal to 80.

82 74 7880

5234 2

234 25 5 80

234 2 400

234 234 2 400 234

To earn a B you must get at least an 83% on the final examination.

CONTINUED

Copyright © 2013, 2009, 2006 Pearson Education, Inc. 1 1 Section 2.7 Solving Linear Inequalities...

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