What you should learn

To solve linear inequalities involving more than one operation

To find the solution set for a linear inequality when replacement values are given for the variables

NCTM Curriculm Standards 2, 6 - 9

 

Introduction: Rosa Whitehair is a partner in an engineering consulting firm. Her fee for consulting on large construction projects is $1000 plus 10% of the design fee that the firm charges its clients. Ms. Whitehair is considering two construction projects: a 50-story office building and the design of an airport terminal. She is interested in both projects but decides to choose the one for which her fee is higher. The company that is designing the office building has agreed to pay Ms. Whitehair a flat fee of $5000. How much does the desing fee need to be for her to choose the terminal?

Let x represent the design fee for the airport terminal.

This inequality involves more than one operation. It can be solved by undoing the operations in reverse of the order of operations in the same way you would solve an equation with more than one operation.

1000 + 0.15x > 5000

1000 - 1000 + 0.15x > 5000 - 1000

0.15x > 4000

(0.15x)/0.15 > 4000/0.15

x > 40,000

If the design fees are higher than $40,000 for the terminal, Ms. Whitehair will choose this project since her consulting fee is higher.

Exercise 1: Determine the value of x so thatA is acute.

ForA to be acute, its measure must be less than 90 degrees

Thus, 3x - 15 < 90

3x - 15 < 90

3x - 15 + 15 < 90 + 15

3x < 105

(3x)/3 < 105/3

x < 35

ForA to be acute, x must be less than 35.

 

 

Sometimes inequalities, like equations, involve variables on each side of the inequality.

Exercise 2: Solve -4w + 9w - 21

When we solve an inequality, the solution set usually includes all numbers for a certain criteria, such as {x|x > 4}. Sometimes a replacement set is given from which the solution set can be chosen.

Exercise 3: Determine the solution set for 3x + 6 > 12 if the replacement set for x is {-2, -1, 0, 1, 2, 3, 4, 5}.

There are two methods in which this can be done.

Method 1

Substitute values into the inequality to find the values that satisfy the inequality. For example, try -2

3x + 6 > 12

3 (-2) + 6 > 12

-6 + 6 > 12

0 > 12 FALSE

Continue this with the rest of the set. Then the true statements are the solution set.

Method 2

Solve the inequality for all values of x. Then determine which values from the replacement set belong to the solution set.

3x + 6 > 12

3x+ 6 - 6 > 12 - 6

3x > 6

x > 2

The solution set is those nubmers from the replacement set that are greater than 2. Thus, the solution set is {3, 4, 5}.

 

 

When solving some inequalities that contain grouping symbols, remember to first use the distributive property to remove the grouping symbols.

Exercise 4: Solve 5(k + 4) - 2(k + 6)5(k + 1) - 1. Then graph the solution.

Activity: Exploration: Graphing Calculators

You can use the inequality symbols in the TEST menu on the graphing calculator to find the solution to an inequality in one variable.

YOUR TURN

a. Clear the Y= list. Enter 3x + 6 > 4x + 9 as Y1. (The symbol > is item 3 on the TEST menu). Press . Describe what you see.

b. Use the TRACE function to scan the values along the graph. What do you notice about the values of y on the graph?

c. Solve the inequalitiy algebraically. How does your solution compare to the pattern you noticed in part b?

Closing Activity: Check for understanding by using this as a quick review before class is over. It should take about the last five to ten minutes. I would use it for my students as their 'ticket out the door'. Click Here.

Homework: The homework to be assigned for tonight would be: 17 - 45 odd, 46, 47, 49, 51 - 60

Alternative Homework: Enriched: 16 - 44 even, 45 - 60

Extra Practice: Students book page 772 Lesson 7-3

Extra Practice Worksheet: Click Here.

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