Module 1 ~ Topic 2Solving Inequalities

Table of Contents

Slides 3-4: Introduction to Inequalities Slides 5-6: Notation and Properties Slide 7: Multiplying/Dividing by a Negative NumberSlides 9-15: Examples Slides 16 – 17: Compound Inequalities Slides 18-20: Interval Notation Slides 23-27: ExamplesSlides 28-30: Graphing Solutions

Audio/Video and Interactive Sites

Slide 2: Graphing Calculator Use Guide Slide 9: Video Slide 19: Interactive SiteSlide 21: VideosSlide 22: Gizmos

Special InstructionsSpecial InstructionsThis module includes graphing calculator work.

Refer to the website, TI-83/84 calculator instructions, for resources and instructions on how to use your calculator to obtain the results we do throughout the lessons. I suggest that you bookmark this site if you haven’t done so already.

Take careful notes and following along with every example in each lesson. I encourage you to ask me questions and think deeply as you’re studying these concepts!

Topic #2: Solving Linear Topic #2: Solving Linear InequalitiesInequalitiesSome real-life problems that

involve mathematics are based in determining a range of values that satisfy the expressions involved.

For these types of problems, we need to set up inequalities, as opposed to equations, to properly solve them.

Solving an Inequality is just like solving an equation,

almost….

Instead of an equal sign there is an inequality sign.

If you multiply or divide by a negative number, you must flip the inequality sign.

There is more than one number that is a part of the solution

There can be 2 inequality signs in the problem.

NotationNotation We will begin by reminding ourselves of formal

mathematical notation and the properties of inequalities.

Greater than > Ex: 12 > 6 Greater than or equal to > Ex: 17 >

16 Less than < Ex: 9 < 13 Less than or equal to < Ex: -15 <

15

PropertiesPropertiesLet a, b, c, and d be real numbers. (The following

apply to all inequality types (> ,< , >, and < ) analogously.)

1. If a < b and b < c then a < c. (Transitive Property)

2. If a < b and c < d then a + c < b + d. (Addition Property)

3. If a < b, then a + c < b + c. (Addition of a Constant)

4. a. If a < b and c is positive, then ac < ab. In other words, the inequality is preserved.

b. If a < b and c is negative, then ac > bc. In other words, the inequality reverses direction. (Multiplication by a Nonzero Constant)

Multiplying by Negatives Multiplying by Negatives Note: Let’s take a moment to understand why, when

we multiply both sides of an inequality by a negative number, the inequality reverses direction.

Well, if we consider the inequality 3 > 1 , how do the two numbers’ opposites relate? In other words, how do -1 and -3 relate? Which one is larger? We know that -1 is larger than -3; i.e. -3 < -1.

Notice that the inequality changes direction, because we are now looking at the left side of zero on the number line.

Definiton:A mathematical inequality is a statement that

one expression is more than, more than or

equal to, less than, or less than or equal to

another. ***Not just simply equal to another

Definiton:

A linear inequality in one variable is an

inequality in the form ax + b < c or ax + b <

c, where a, b, and c are real numbers.

Example:Example:Solve the inequality: 4x + 6 < 10

Solution: What does solving this inequality? It means that we are looking for the values of x that would make the value of 4x + 6 be less than 10. Can you think of some?

So, let’s find all of our solutions at once. We solve inequalities in much the same way as we do equations.

What about 0, -18, 9, 3.6, -2.034, -1000, 2,

…there’s a lot…too many to list

Watch this video for a quick refresher!!!

4 6 10

4 4

4 4

6 6

4 4

1

x

x

x

x

This means that any value for x less than 1 would make the inequality 4x + 6 < 10 a true statement.

For example, if x = -3 we have 4(-3)+6 = -6 < 10

However, if x = 1 we have 4(1) + 6 = 10 < 10. This is False ~ 10 is not less than 10!

So we can only have answers that are less than 1 to make this a true statement.

Think about what numbers are solutions to this problem.

Solve the inequality: 4x + 6 < 10

(Any number less than 1.)

How many solutions are there to this inequality? (An infinite amount)

Solve the inequality: -9 < x + 6 < 10

x

x

15

69

4

106

x

x

1069 x

This inequality can be solved 2 different ways, but the answer will be the same wither way.

415 x

415 x

1069 x

415 x

6 6 6

More Examples:More Examples:a) Solve: 3x – 10 > 9x + 5

Solution: 2

5x

b) Solve: -7 < 5x – 3 < 2 This is a compound inequality *****Answer will also be a compound inequality

Solution: 15

4 x

x

x

x

x

xx

2

56

15

615

5610

59103

15

4

554

2357

x

x

x

c) Suppose you were working in the finance department of a factory that produces widgets. The total cost of producing widgets depends on the number of them produced. In order to make one batch, your factory has a fixed (one-time) cost of $1500, and it costs $35 to produce each widget.

1. Write an equation that represents the total cost, C, of producing w widgets.

2. Suppose the factory has a budget of at most $32,000 to spend on making a batch of widgets. What is the range of the number of widgets your factory can produce?

Solutions:

1. If C is the total cost of producing a batch, and w is the number of widgets, then the equation that represents their relationship is C = 35w + 1500

2. Since the budget is at most $32,000, we want to find the range of values for w such that 35w + 1500 < 32000. We include 32,000 in our inequality, since the budget includes 32,000 as a possibility. We then proceed to solve the inequality.

So, 6100/7 is approximately 871.43. Since w needs to be less than or equal to 6100/7, and it MUST be a whole number (you wouldn’t want a fraction of a widget) we will have to round to 871.

So, the answer to our finance question is that our factory can produce at most 871 widgets to meet the budget. (We could have also said this by saying that our factory can produce no more than 871 widgets to meet the budget.)

1500 15

35 1500 32000

35 30500

35 30500

00

35 5

00

3

61

7

w

w

w

w

d) Suppose you’re making a slow-cooking stew, and the recipe states that the temperature of your stew must remain strictly between 45ºC and 55ºC for 3 hours in order to cook properly. What would this range be in degrees Fahrenheit?

Solution: The relationship between Celsius and Fahrenheit is

Since we want to find F such that 45 < C < 55, the inequality we want to solve

is

So, we find that the stew must remain strictly between 113ºF and 131ºF for 3 hours to cook properly.

532

9FC

545 32 55

9F

545 32 55

9

545 32 55

9

81 32 99

8

9 9 9

5 5 5

32 32 3

1 32 99

11

2

3 131

F

F

F

F

F

You are in college. Your final math class grade consists of 4 test grades. You’ve earned a 78%, 88%, and 92%, respectively, on your first three tests this semester. Your parents tell you must get a final grade between 82% and 89%.

What range of scores could you earn on your fourth test in order to have a class average between 82% and 89%?

You know that to find the average of a set of numbers, you add them and divided by the number of items you added.

You know 3 of the 4 test scores here. For any unknown, we can call it x.

We want our answer to be a range of scores, all the scores, that give us the average we are looking for,

The lowest average we want is 82% and the highest average is 89%

4

928878 yettakennothaveyoutestforscoretest

4

928878 x

894

92887882

x

894

25882

x

Go to next slide to finish problem

894

25882

x

356258328 x

9870 x

)89(44

258)4()82(4

x

-258 -258 -258

This means that the lowest score you can get on your 4th test is a 70% and the high score of 98% on the 4th test will give you an average of an 89%. Any score between the 70% and 98% will give you an average between 82% and 89%

Interval NotationThere is another way to write the solutions to linear inequalities. It is called interval notation. Here is a summary:

< or > (without the equal sign) is shown by using (< or > is shown by using [ (infinity) is used when there is a single sided

inequality. So, in the example above 113 < F < 131, we would write in interval notation as [113,131).

More examples: 5< x < 9 is written as : (5, 9] 6 < x < 25 is written as: [6, 25) Now, a single sided inequality uses the infinity symbol: x > 5 is written as : (5, ). We use the parenthesis at 5 because there is no equals sign under the greater than sign. The infinity symbol always uses the parenthesis. x < 15 is written as : ( , 15].

Interval NotationWrite the following solution in interval notation:Write the following solution in interval notation:

Notice the solid circle gets the [ because -1 is included, and the open circle gets the ) because 2 is not included.

Interactive Example of Interval Notation and Set

Notation(Double-click on endpoints to change between included and

excluded endpoints)

Interval NotationWrite the following solutions in interval notation:Write the following solutions in interval notation:

3.9

40

8

9

3

4

3

4.63.1

132

x

x

x

x

x

x

),3.9[

4,0

8

9,

3

4

)3,(

]4.6,3.1[

]13,2(

More on Interval Notation

Interval Notation

Examples, Interval Notation,

Graphing on Number Line

Practice Problems and Answers

How to Solve Multistep

Inequalities

How to Solve Inequalities and

Graph on a Number Line

Compound Inequalities

Solving and Graphing

Compound Inequalities

Gizmos Gizmos

Gizmo: Solving Inequalities using

Multiplication and Division

Gizmo: Compound Inequalities

Gizmo: Solving Inequalities using

Addition and Subtraction

Gizmo: Solving Inequalities using

Multiplication and Division

Example 1:Example 1:10

1

9

1

2

1

3

5 mm

10

190

9

190

2

190

3

590 mm

10

1

9

190

2

1

3

590 mm

91045150 mm-45 -45

5410150 mm-10m -10m

54140 m140 140

70

27140

54

m

m

Remember this problem?

We did this as an equation in Topic 1 Notes.

Nothing changes, except the sign.

Practice ExamplesPractice Examples

Example 2:Example 2:

Example 3:Example 3:

Solve the equation 3p + 2 < 0 .

Solve the equation -7m – 1 > 0.

Solutions on next slide. Solve these on your own first.

Example 4:Example 4:Solve the equation 14z – 28 > 0.

Practice Examples AnswersPractice Examples AnswersExample 2:

Example 3:

Solve the equation 3p + 2 < 0 .

Solve the equation -7m – 1 > 0.

Example 4:Solve the equation 14z – 28 > 0.

3

2p

7

1m

2z

Notice the sign flips because you divide by a – 7.

More Practice ExamplesMore Practice ExamplesExample 5:

Example 6:

Solve the equation .

Solve the equation .

Solve these on your own first. Solutions on next slide.

19684 nn

542

1

3 x

x

More Practice Examples - More Practice Examples - AnswersAnswers

Example 5:

Example 6:

Solve the equation .

Solve the equation .

19684 nn

542

1

3 x

x

2

27

272

6 6

2764

8 8

19684

n

n

nn

nn

nn

2

27n

26

123

12326

120326

1202432

20462

1

36

2042

1

3

542

1

3

x

x

x

xx

xx

xx

xx

26

123x

Graphing Answers to Graphing Answers to InequalitiesInequalities

When graphing and inequality:

< shade left (toward negatives) < shade left (toward negatives)

> shade right (toward positives) > shade right (toward positives)

When graphing and inequality:

< open circle > open circle

< closed circle > closed circle

When the variable is to the left of the inequality sign:

Lets graph some of the Lets graph some of the answers we have already answers we have already

found:found:

Interval Notatio

n

1xSlide 10

415 x

Slide 11

2

5x

Slide 12

Slide 17

9870 x

Slide 25

2z

1,

4,15

2

5,

1,

5

4

98,70

,2

15

4 x

0 1

0 4-15

0

0

0

0

- 5/2

1- 4/5

70 98

2