Transcript Module1Topic2Notes

Module 1 ~ Topic 2
Solving Inequalities
Table of Contents
 Slides 3-4: Introduction to Inequalities
 Slides 5-6: Notation and Properties
 Slide 7: Multiplying/Dividing by a Negative Number
Slides 9-15: Examples
 Slides 16 – 17: Compound Inequalities
 Slides 18-20: Interval Notation
Slides 23-27: Examples
Slides 28-30: Graphing Solutions
Audio/Video and Interactive Sites
Slide 2: Graphing Calculator Use Guide
Slide 9: Video
Slide 19: Interactive Site
Slide 21: Videos
Slide 22: Gizmos
Special Instructions

This 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 Inequalities
Some 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.
Notation

We will begin by reminding ourselves of formal mathematical
notation and the properties of inequalities.

Greater than
Greater than or equal to
Less than
Less than or equal to



>
>
<
<
Ex:
Ex:
Ex:
Ex:
12 > 6
17 > 16
9 < 13
-15 < 15
Properties

Let 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 

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.
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?
What about
0, -18, 9, 3.6, -2.034, -1000, 2,
…there’s a lot…too many to list
So, let’s find all of our solutions at once. We solve inequalities in much
the same way as we do equations.
Watch this video for a quick refresher!!!
Solve the inequality: 4x + 6 < 10
4 x  6  10
6 6
This means that any value for x less than 1 would
make the inequality 4x + 6 < 10 a true statement.
4x  4
For example, if x = -3 we have 4(-3)+6 = -6 < 10
4x
However, if x = 1 we have 4(1) + 6 = 10 < 10.
This is False ~ 10 is not less than 10!
4

4
4
x 1
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.
(Any number less than 1.)
How many solutions are there to this inequality?
(An infinite amount)
Solve the inequality: -9 < x + 6 < 10
This inequality can be solved 2 different ways, but the answer
will be the same wither way.
9  x  6  10
9  x  6  10
6
9  x  6
 15  x
x  6  10
x4
15  x  4
15  x  4
6 6
15  x  4
More Examples:
3 x  10  9 x  5
a) Solve: 3x – 10 > 9x + 5
Solution: x  
5
2
 10  6 x  5
 15  6 x
15
x
6
5
 x
2

b) Solve: -7 < 5x – 3 < 2
This is a compound inequality
*****Answer will also be a compound inequality
Solution: 
4
 x 1
5
7  5 x  3  2
 4  5x  5
4
  x 1
5
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.
35w  1500  32000
 1500  1500
35w  30500
35w
35
w 

30500
35
6100
7
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.)
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:
5
The relationship between Celsius and Fahrenheit is C   F  32 
9
Since we want to find F such that 45 < C < 55, the inequality we want to solve
is 45 
5
9
 F  32   55
45 
5
9
 F  32   55
 9
  F  32     55 
5
59
 5
81  F  32  99
81  F  32  99
9
 45  
32
95
 32  32
113  F  131
So, we find that the stew must remain strictly between 113ºF and 131ºF for 3 hours
to cook properly.
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%
78  88  92  test score for test you have not taken yet
4
78  88  92  x
4
82 
78  88  92  x
 89
4
258  x
82 
 89
4
Go to next slide to finish problem
258  x
82 
 89
4
258  x
4(82)  (4) 
 4(89)
4
328  258  x  356
-258
-258
-258
70  x  98
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 Notation
There 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 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 Notation
Write the following solutions in interval notation:
2  x  13
( 2,13]
 1 .3  x  6 .4
[ 1.3,6.4]
x  3
(,3)
4
9
 x
3
8
 4 9
 3 , 8 


0 x4
0,4 
x  9 .3
[ 9.3,  )
How to Solve
Inequalities and
Graph on a Number
Line
How to Solve
Multistep Inequalities
Compound
Inequalities
Solving and Graphing
Compound
Inequalities
Interval Notation
More on Interval
Notation
Examples, Interval
Notation, Graphing
on Number Line
Practice Problems and
Answers
Gizmos
Gizmo: Solving Inequalities using
Addition and Subtraction
Gizmo: Solving Inequalities using
Multiplication and Division
Gizmo: Solving Inequalities using
Multiplication and Division
Gizmo: Compound Inequalities
Example 1:
5
1
1
1
m

m
3
2
9
10
1
1 
5
1
90 m    90 m  
2
10 
3
9
5 
1
1 
 1 
90 m   90   90 m   90 
3 
2
9 
 10 
150m  45  10m  9
-45
-45
150m  10m  54
-10m -10m
140m  54
140
140
54
140
27
m
70
m
Remember this
problem?
We did this as an
equation in Topic 1
Notes.
Nothing changes,
except the sign.
Practice Examples
Example 2: Solve the equation 3p + 2 < 0 .
Example 3: Solve the equation -7m – 1 > 0.
Example 4: Solve the equation 14z – 28 > 0.
Solutions on next slide. Solve these on your own first.
Practice Examples Answers
Example 2: Solve the equation 3p + 2 < 0 .
p
2
3
Example 3: Solve the equation -7m – 1 > 0.
m
1
7
Notice the sign flips because you divide by a – 7.
Example 4: Solve the equation 14z – 28 > 0.
z2
More Practice Examples
Example 5: Solve the equation
4n  8  6n  19 .
x 1
  4 x  5.
Example 6: Solve the equation
3 2
Solve these on your own first. Solutions on next slide.
More Practice Examples - Answers
Example 5: Solve the equation
4n  8  6n  19 .
27
n
2
x 1
  4 x  5.
Example 6: Solve the equation
3 2
123
x
26
4n  8  6n  19
8
8
 4n  6n  27
 6n  6n
2n  27
n
27
2
x 1
  4 x  5
3 2
x 1
  4 x  20
3 2
 x 1
6    6 4 x  20
3 2
2 x  3  24 x  120
26 x  3  120
26 x  123
123
x
26
Graphing Answers to Inequalities
When the variable is to the left of the inequality sign:
When graphing and inequality:
When graphing and inequality:
< open circle
> open circle
<
<
< closed circle
> closed circle
> shade right (toward positives)
> shade right (toward positives)
shade left (toward negatives)
shade left (toward negatives)
Lets graph some of the answers we
have already found:
Interval
Notation
Slide 10
x 1
,1
15  x  4
15,4 
0 1
-15
0 4
Slide 11
x
5
2
Slide 12
4
  x 1
5
5

  ,  
2

0
- 5/2
 4 
 5 ,1


- 4/5
0 1
Slide 17
70  x  98
Slide 25
z2
70,98
2,  
0
70
0
2
98