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Week 2
Due for this week…




Homework 2 (on MyMathLab – via the Materials
Link)  Monday night at 6pm.
Read Chapter 3 and 8.1, 8.2
Do the MyMathLab Self-Check for week 2.
Learning team planning for week 5.
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 2
2.1
Introduction to Equations
Basic Concepts
The Addition Property of Equality
The Multiplication Property of Equality
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
The Linear Equation

It has one variable : x
ax + b = 0
Where a and b are real numbers and a is not equal to
0.
A photo album of linear equations







2x+3=0
x+8=0
3x=7
2x+5=9-5x
3+5(x-1)=-7+x
3=8x+5
Etc.
(say cheese!)
How to legally play with linear equations.

Most of the magic we work with are really obvious
tricks.
 Remember what happens if we multiply by things
that equal 1? e.g. 5/5 10/10 ?

Adding the SAME THING to both sides of an
equation does not change the equation…
The Addition Property of Equality



So if you start with a = b
You don’t change anything if you take a new
number or variable and add it to both sides!
a+c = b+c
The GOAL

We want x all by itself on one side and the rest of
the junk on the other (so we can use our calculator
or fingers).

The technical term for this is: solving for x
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 9
EXAMPLE
Using the addition property of equality
Solve each equation.
a. x + 21 = 9
b. n – 8 = 17
Solution
a.
x + 21 = 9
x + 21 + (−21) = 9 + (−21)
b.
n – 8 = 17
n – 8 + 8 = 17 + 8
x + 0 = −12
n + 0 = 25
x = −12
n = 25
The solution is −12.
The solution is 25.
** Try exercises 17-26
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 10
EXAMPLE
Solving and checking a solution
Solve the equation −7 + x = 12 and then check the
solution.
Solution
Isolate x by adding 7 to each side.
−7 + x = 12
−7 + 7 + x = 12 + 7
0 + x = 19
x = 19
The solution is 19.
Check:
−7 + x = 12
?
−7 + 19 = 12
12 = 12
The answer checks.
** Try exercises 27-34
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 11
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 12
EXAMPLE
Using the multiplication property of equality
Solve each equation.
1
a.
x 8
6
b. −11a = 33
Solution
a.
1
x 8
6
1
6 x  86
6
1 x  48
x  48
The solution is 48.
b.
−11a = 33
11a 33

11 11
a  3
The solution is −3.
** Try exercises 41-48
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 13
EXAMPLE
Solving and checking a solution
2
2
Solve the equation   x and then check the
5
9
solution.
Solution
2
2
 x
5
9
9 2
2  9
       x
2 5
9  2
9
  1 x
5
9
 x
5
9
The solution is  .
5
** Try exercises 49-58
Check:
2
2
 x
5
9
2 ? 2 9
   
5
9 5
2 2

5 5
The answer checks.
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley-
Slide 14
EXAMPLE
Application
A veterinary assistant holds a cat and steps on a
scale. The scale reads 153 lbs. Alone the assistant
weighs 146 lbs.
a. Write a formula to show the relationship of the
weight of the cat, x, and the assistant.
b. Determine how much the cat weighs.
Solution
a. 146 + x = 153
b.
146 + x = 153
146 + (−146) + x = 153 − 146
x=7
The cat weighs 7 lbs.
** Try exercises 61-70
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 15
2.2
Linear Equations
Basic Concepts
Solving Linear Equations
Applying the Distributive Property
Clearing Fractions and Decimals
Equations with No Solutions or Infinitely Many
Solutions
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
If an equation is linear, writing it in the form ax + b = 0
should not require any properties or processes other than
the following:
• using the distributive property to clear parentheses,
• combining like terms,
• applying the addition property of equality.
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 17
EXAMPLE
Determine whether an equation is linear
Determine whether the equation is linear. If the
equation is linear, give values for a and b.
a. 9x + 7 = 0
Solution
b.
5x3
+9=0
5
c. 7   0
x
a. The equation is linear because it is in the form
ax + b = 0 with a = 9 and b = 7.
b. This equation is not linear because it cannot be
written in the form ax + b = 0. The variable has an
exponent other than 1.
c. This equation is not linear because it cannot be
written in the form ax + b = 0. The variable appears in
the denominator of a fraction.
** Try exercises 13-26
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 18
EXAMPLE
Using a table to solve an equation
Complete the table for the given values of x. Then
solve the equation 4x – 6 = −2.
x
−3
−2
4x − 6
−18
−1
0
1
2
3
Solution
−3
x
4x − 6
−2
−1
0
1
2
3
−18 −14 −10 −6
−2
2
6
From the table 4x – 6 = −2 when x = 1. Thus the
solution to 4x – 6 = −2 is 1.
** Try exercises 27-30
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 19
Slide 20
Rules for Happiness



First do whatever adding and subtracting you can
do…
THEN do whatever multiplication and division you
can do.
THEN you should be done!
EXAMPLE
Solving linear equations
Solve each linear equation.
a. 12x − 15 = 0
b. 3x + 19 = 5x + 5
Solution
a.
12x − 15 = 0
12x − 15 + 15 = 0 + 15
12x = 15
12 x 15

12 12
15 5
x

12 4
** Try exercises 39-42
3x + 19 = 5x + 5
b.
3x − 3x + 19 = 5x − 3x + 5
19 = 2x + 5
19 − 5 = 2x + 5 − 5
14 = 2x
14 2x

2
2
7x
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 22
EXAMPLE
Applying the distributive property
Solve the linear equation. Check your solution.
x + 5 (x – 1) = 11
Solution
x + 5 (x – 1) = 11
x + 5 x – 5 = 11
6x − 5 = 11
6x − 5 + 5 = 11 + 5
6x = 16
6 x 16

6
6
16
x
6
8

3
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 23
EXAMPLE
Check
continued
x + 5 (x – 1) = 11
8
8 
 5   1  11
3
3 
33
 11
3
8
8 3
 5     11
3
3 3
8
5
 5    11
3
3
8 25

 11
3 3
11  11
The answer checks, the
8
solution is .
3
** Try exercises 43-52
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 24
EXAMPLE
Clearing fractions from linear equations
Solve the linear equation.
1
1
x x 5
2
6
Solution
1
1
x x 5
2
6
1 
1
6 x  x   56
6 
2
3x  x  30
2x  30
x  15
The solution is 15.
** Try exercises 55-58
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 25
EXAMPLE
Clearing decimals from linear equations
Solve the linear equation.
5.3  0.8x  7
Solution
5.3  0.8x  7
10  5.3  0.8x   7 10
53  8x  70
53   53  8x  70  53
8x  17
8 x 17

8 8
** Try exercises 53-54
17
x
8
17
The solution is  .
8
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 26
Equations with No Solutions or Infinitely Many
Solutions
An equation that is always true is called an identity
and an equation that is always false is called a
contradiction.
Slide 27
EXAMPLE
Determining numbers of solutions
Determine whether the equation has no solutions, one
solution, or infinitely many solutions.
a. 10 – 8x = 2(5 – 4x)
b. 7x = 9x + 2(12 – x)
c. 6x = 4(x + 5)
Solution
a. 10 – 8x = 2(5 – 4x)
10 – 8x = 10 – 8x
– 8x = – 8x
Because the equation 0 = 0 is
always true, it is an identity and
there are infinitely many
solutions.
0=0
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 28
EXAMPLE
continued
b. 7x = 9x + 2(12 – x)
7x = 9x + 24 – 2x
7x = 7x + 24
c. 6x = 4(x + 5)
6x = 4x + 20
2x = 20
x = 10
0 = 24
Because the equation 0 = 24
is always false, it is a
contradiction and there are
no solutions.
Thus there is one
solution.
** Try exercises 63-72
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 29
Slide 30
2.3
Introduction to Problem Solving
Steps for Solving a Problem
Percent Problems
Distance Problems
Other Types of Problems
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 32
EXAMPLE
Translating sentences into equations
Translate the sentence into an equation using the
variable x. Then solve the resulting equation.
a. Six times a number plus 7 is equal to 25.
b. The sum of one-third of a number and 9 is 18.
c. Twenty is 8 less than twice a number.
Solution
a. 6x + 7 = 25
6x = 18
6 x 18

6
6
x 3
b. 1 x  9  18
3
1
x9
3
x  27
c. 20 = 2x − 8
28 = 2x
** Try exercises 11-18
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
28 2 x

2
2
14 = x
Slide 33
EXAMPLE
Solving a number problem
The sum of three consecutive integers is 126. Find the
three numbers.
Solution
Step 1: Assign a variable to an unknown quantity.
n: smallest of the three integers
n + 1: next integer
n + 2: largest integer
Step 2: Write an equation that relates these unknown
quantities.
n + (n + 1) + (n + 2) = 126
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 34
EXAMPLE
continued
Step 3: Solve the equation in Step 2.
n + (n + 1) + (n + 2) = 126
(n + n + n) + (1 + 2) = 126
3n + 3 = 126
3n = 123
n = 41
So the numbers are 41, 42, and 43.
Step 4: Check your answer. The sum of these integers
is 41 + 42 + 43 = 126.
** Try exercises 19-28 The answer checks.
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 35
Note: To write x% as a decimal number, move the
decimal point in the number x two places to the left and
then remove the % symbol.
Slide 36
EXAMPLE
Converting percent notation
Convert each percentage to fraction and decimal notation.
a. 47%
b. 9.8%
c. 0.9%
Solution
47
.
100
Decimal Notation: 47%  0.47.
9.8
98
49  2
49
b. Fraction Notation: 9.8% 



.
100 1000 500  2 500
Decimal Notation: 9.8%  0.098.
0.9
9
c. Fraction Notation: 0.9% 

100 1000
** Try exercises 35-42
Decimal Notation: 0.9%  0.009.
a. Fraction Notation: 47% 
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 37
EXAMPLE
Converting to percent notation
Convert each real number to a percentage.
a. 0.761
2
b.
5
c. 6.3
Solution
a. Move the decimal point two places to the right and
then insert the % symbol to obtain 0.761 = 76.1%
2
2
b.
 0.40, so  40%.
5
5
c. Move the decimal point two places to the right and
then insert the % symbol to obtain 6.3 = 630%. Note
that percentages can be greater than 100%.
** Try exercises 43-54
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 38
EXAMPLE
Calculating a percent increase
The price of an oil change for an automobile increased
from $15 to $24. Calculate the percent increase.
Solution
new value - old value
100
old value
24 - 15

100  60%
15
** Try exercises 55-56
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 39
EXAMPLE
Solving a percent problem
A car salesman sells a total of 85 cars in the first and
second quarter of the year. In the second quarter, he
had an increase of 240% over the previous quarter.
How many cars did the salesman sell in the first
quarter?
Solution
Step 1: Assign a variable.
x: the amount sold in the first quarter.
Step 2: Write an equation.
x + 2.4x = 85
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 40
EXAMPLE
continued
Step 3: Solve the equation in Step 2.
x + 2.4x = 85
3.4x = 85
x = 25
In the first quarter the salesman sold 25 cars.
Step 4: Check your answer.
An increase of 240% of 25 is 2.4 × 25 = 60.
Thus the amount of cars sold in the second
quarter would be 25 + 60 = 85.
** Try exercises 57-60
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 41
EXAMPLE
Solving a distance problem
A truck driver travels for 4 hours and 30 minutes at a
constant speed and travels 252 miles. Find the speed
of the truck in miles per hour.
Solution
Step 1: Let r represent the truck’s rate, or speed, in
miles.
Step 2: The rate is to be given in miles per hour, so
change the 4 hours and 30 minutes to 4.5 or 9/2
hours.
d = rt
9
252   r
2
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 42
EXAMPLE
continued
Step 3: Solve the equation.
9
252   r
2
2
2 9
 252    r
9
9 2
56  r
The speed of the truck is
56 miles per hour.
Step 4:
d = rt  56 
9
 252 miles
2
The answer checks.
** Try exercises 71-74
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 43
EXAMPLE
Mixing chemicals
A chemist mixes 100 mL of a 28% solution of alcohol
with another sample of 40% alcohol solution to obtain
a sample containing 36% alcohol. How much of the
40% alcohol was used?
Solution
Step 1: Assign a variable.
x: milliliters of 40%
x + 100: milliliters of 36%
Step 2: Write an
equation.
Concentration
Solution Amount
(milliliters)
Pure alcohol
0.28
100
28
0.40
x
0.4x
0.36
x + 100
0.36x + 36
0.28(100) + 0.4x = 0.36(x + 100)
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 44
EXAMPLE
continued
Step 3: Solve the equation in Step 2.
0.28(100) + 0.4x = 0.36(x + 100)
28(100) + 40x = 36(x + 100)
2800 + 40x = 36x + 3600
2800 + 4x = 3600
4x = 800
x = 200
200 mL of 40% alcohol solution was added to the
100 mL of the 28% solution.
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 45
EXAMPLE
continued
Step 4: Check your answer.
If 200 mL of 40% solution are added to the 100 mL
of 28% solution, there will be 300 mL of solution.
200(0.4) + 100(0.28) = 80 + 28 = 108 of pure alcohol.
The concentration is
108
 0.36, or 36%.
300
** Try exercises 79-80
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 46
2.4
Formulas
Basic Concepts
Formulas from Geometry
Solving for a Variable
Other Formulas
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE
Calculating mileage of a trip
A tourist starts a trip with a full tank of gas and an
odometer that reads 59,478 miles. At the end of the
trip, it takes 8.6 gallons of gas to fill the tank, and the
odometer reads 59,715 miles. Find the gas mileage
for the car.
Solution
The distance traveled is 59,715 – 59, 478 = 237 miles
and the number of gallons used is G = 8.6. Thus,
D
237
 27.6 miles per gallon.
M

G
8.6
** Try exercises 81-82
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 48
EXAMPLE
Calculating area of a region
A residential lot is shown.
Find the area of this lot.
205 ft
372 ft
Solution
116 ft
The area of the rectangle: AR  LW
AR  372  205
AR  76, 260 square feet
The area of the triangle: AT  12 bh
AT  12 116  372
AT  21,576 square feet
Total area = 76,260 + 21,576 = 97,836 square feet.
** Try exercises 19-29
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 49
EXAMPLE
Finding angles in a triangle
In a triangle, the smaller angles are equal in measure
and are one-third of the largest angle. Find the
measure of each angle.
Solution
Let x represent the measure of each of the two smaller
angles. Then the measure of the largest angle is 3x, and
the sum of the measures of the three angles is given by
x  x  3x  180
5 x  180
5 x 180

5
5
x  36
The measure of the largest
angle is 3x, thus 36 ∙ 3 = 108°.
The measure of the three
angles are 36°, 36°, and 108°.
** Try exercises 31-38
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 50
EXAMPLE
Finding the volume and surface area of a box
Find the volume and the
surface area of the box shown.
6 cm
5 cm
Solution
12 cm
The volume of the box is
V = LHW
V = 12 ∙ 6 ∙ 5
V = 360 cm3
The surface area of the box is
S  2LW  2WH  2LH
S  2(12)(5)  2(5)(6)  2(12)(6)
S  120  60  144
** Try exercises 41-46 S  324 square centimeters
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 51
EXAMPLE
Calculating the volume of a soup can
A cylindrical soup can has a
radius of 2 ½ inches and a
height of 5 85 inches. Find
the volume of the can.
∙
r
h
Solution
V   r 2h
2
 5   45 
V     
2  8 
 1125 
V  

 32 
V  110.45 cubic inches
** Try exercises 51
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 52
EXAMPLE
Solving for a variable
Solve each equation for the indicated variable.
yz
a. 3 x 
for z
5
b. np  nm  nq for p
Solution
yz
5
15 x  y  z
a. 3 x 
15 x  y  z
z  15 x  y
b. np  nm  nq for p
np  nq  nm
np  n(q  m)
n( q  m)
p
n
p  qm
** Try exercises 53-64
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 53
Other Formulas
To calculate a student’s GPA, the number of credits
earned with a grade of A, B, C, D, and F must be
known. If a, b, c, d, and f represent these credit
counts respectively, then
4a  3b  2c  d
GPA 
.
abcd  f
Slide 54
EXAMPLE
Calculating a student’s GPA
A student has earned 18 credits of A, 22 credits of B, 8
credits of C and 4 credits of D. Calculate the student’s
GPA to the nearest hundredth.
Solution
Let a = 18, b = 22, c = 8, d = 4 and f = 0
4 18  3  22  2  8  4 158
 3.04
GPA 

18  22  8  4  0
52
The student’s GPA is 3.04.
** Try exercises 69-72
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 55
EXAMPLE
Converting temperature
The formula C  95  F  32 is used to convert
degrees Fahrenheit to degrees Celsius.
Use this formula to convert 23°F to an equivalent
Celsius temperature.
Solution
5
C   F  32 
9
5
C   23  32 
9
5
C   9  = −5°C
9
** Try exercises 73-80
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 56
2.5
Linear Inequalities
Solutions and Number Line Graphs
The Addition Property of Inequalities
The Multiplication Property of Inequalities
Applications
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Solutions and Number Line Graphs
A linear inequality results whenever the equals sign
in a linear equation is replaced with any one of the
symbols <, ≤, >, or ≥.
x > 5,
3x + 4 < 0, 1 – y ≥ 9
A solution to an inequality is a value of the variable
that makes the statement true. The set of all solutions
is called the solution set.
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 58
EXAMPLE
Graphing inequalities on a number line
Use a number line to graph the solution set to each
inequality.
a. x  1
b. x  1
c. x  5
d. x  2
** Try exercises 13-20
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 59
Interval Notation
Each number line graphed on the previous slide
represents an interval of real numbers that corresponds
to the solution set to an inequality.
Brackets and parentheses can be used to represent the
interval.
For example:
x  1  (1, )
x  1  [1, )
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 60
EXAMPLE
Writing solution sets in interval notation
Write the solution set to each inequality in interval
notation.
a. x  6
b. y  2
(6, )
(, 2]
** Try exercises 27-32
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 61
EXAMPLE
Checking possible solutions
Determine whether the given value of x is a solution to
the inequality. 4 x  2  8,
x7
?
4 x  2 <8
?
4(7)  2  8
?
28  2  8
?
26  8 False
** Try exercises 33-42
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 62
The Addition Property of Inequalities
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 63
EXAMPLE
Applying the addition property of inequalities
Solve each inequality. Then graph the solution set.
a. x – 2 > 3
b. 4 + 2x ≤ 6 + x
Solution
a.
x–2>3
x–2+2>3+2
x>5
b. 4 + 2x ≤ 6 + x
4 + 2x – x ≤ 6 + x – x
4+x≤6
4–4+x≤6–4
x≤2
** Try exercises 51-58
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 64
The Multiplication Property of Inequalities
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 65
Applying the multiplication property of
inequalities
EXAMPLE
Solve each inequality. Then graph the solution set.
1
a. 4x > 12
b. 2   x
4
Solution
a.
4x > 12
4 x 12

4
4
x 3
1
b. 2   x
4
1
4(2)  (4)  x
4
8 x
** Try exercises 59-66
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 66
EXAMPLE
Applying both properties of inequalities
Solve each inequality. Write the solution set in
set-builder notation.
a. 4x – 8 > 12
b. 4  3x  4x  5
Solution
a.
4x – 8 > 12
4x  8  8  12  8
4x  20
x 5
{x | x  5}
b. 4  3x  4x  5
4  3x  3x  4x  5  3x
4  x  5
4  5  x  5  5
9  x
{x | x  9}
** Try exercises 71-100
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 67
Applications
To solve applications involving inequalities, we often
have to translate words to mathematical statements.
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 68
EXAMPLE
Translating words to inequalities
Translate each phrase to an inequality. Let the variable
be x.
a. A number that is more than 25.
x > 25
b. A height that is at least 42 inches.
x ≥ 42
** Try exercises 101-108
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 69
EXAMPLE
Calculating revenue, cost, and profit
For a snack food company, the cost to produce one
case of snacks is $135 plus a one-time fixed cost of
$175,000 for research and development. The revenue
received from selling one case of snacks is $250.
a. Write a formula that gives the cost C of producing x
cases of snacks. C = 135x + 175,000
b. Write a formula that gives the revenue R from selling
x cases of snacks.
R = 250x
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 70
EXAMPLE
Calculating revenue, cost, and profit
For a snack food company, the cost to produce one
case of snacks is $135 plus a one-time fixed cost of
$175,000 for research and development. The revenue
received from selling one case of snacks is $250.
c. Profit equals revenue minus cost. Write a formula
that calculates the profit P from selling x cases of
snacks.
P=R–C
= 250x – (135x + 175,000)
= 115x – 175,000
d. How many cases need to be sold to yield a positive
profit?
115x – 175,000 > 0
115x > 175,000
x > 1521.74
Must sell at least 1522 cases.
** Try exercises 119-120
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 71
End of week 2
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You again have the answers to those problems not
assigned
Practice is SOOO important in this course.
Work as much as you can with MyMathLab, the
materials in the text, and on my Webpage.
Do everything you can scrape time up for, first the
hardest topics then the easiest.
You are building a skill like typing, skiing, playing a
game, solving puzzles.
NEXT TIME: Linear Equations w/2 variables and
Graphing + slope and y-intercepts