Transcript bei05_ppt_02

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Slide 1- 1
2
2.1
2.2
2.3
2.4
2.5
2.6
2.7
Equations, Inequalities, and
Problem Solving
Solving Equations
Using the Principles Together
Formulas
Applications with Percent
Problem Solving
Solving Inequalities
Solving Applications with Inequalities
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2.1
Solving Equations

Equations and Solutions

The Addition Principle

The Multiplication Principle

Selecting the Correct Approach
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Solution of an Equation
Any replacement for the variable that
makes an equation true is called a solution
of the equation. To solve an equation
means to find all of its solutions.
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Slide 2- 4
Example
Determine whether 8 is a solution of x + 12 = 21.
Solution
x + 12 = 21
8 + 12 | 21
20  21
Writing the equation
Substituting 8 for x
False
Since the left-hand and right-hand sides differ, 8
is not a solution.
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Slide 2- 5
Equivalent Equations
Equations with the same solutions are called equivalent
equations.
The Addition Principle
For any real numbers a, b, and c,
a = b is equivalent to a + c = b + c.
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Slide 2- 6
Example
Solve: 8.3 = y  17.9
Solution
8.3 = y  17.9
8.3 + 17.9 = y  17.9 + 17.9
9.6 = y
Check: 8.3 = y  17.9
8.3 | 9.6  17.9
8.3 = 8.3
The solution is 9.6.
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Slide 2- 7
The Multiplication Principle
For any real numbers a, b, and c with c  0,
a = b is equivalent to a • c = b • c.
Example
3
Solve: x  15
4
3
x  15
Solution
4
4 3
4
 x  15 
3 4
3
Multiplying both sides by 4/3.
1x  20
x  20
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Slide 2- 8
Example
Solve: 7x = 84
Solution
7x = 84
7 x 84

7 7
1 x  12
x  12
Check:
Dividing both sides by 7.
7x = 84
7(12) | 84
84 = 84
The solution is –12.
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Slide 2- 9
2.2 Using the Principles Together

Applying Both Principles

Combining Like Terms

Clearing Fractions and Decimals

Contradictions and Identities
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Example
Solve: 9 + 8x = 33
Solution
9 + 8x = 33
9 + 8x  9 = 33  9 Subtracting 9 from both sides
9 + ( 9) + 8x = 24
8x = 24
8 x 24
Dividing both sides by 8

8
8
x=3
Check: 9 + 8x = 33
9 + 8(3) | 33
9 + 24 | 33
33 = 33
The solution is 3.
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Example
Solution
2
Solve: x  5  11
3
2
x  5  11
3
2
x  5  5  11  5
3
2
x  16
3
3 2
3
 x  16 
2 3
2
8 2 3
1 x 
2
Adding 5 to both sides
Multiplying both sides by 3/2.
Simplifying
x  24
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Slide 2- 12
Example Solve: 36  t = 17
Solution
36  t = 17
36 from both
36  t  36 = 17  36 Subtracting
sides
t = 19
(1)  t = (19)(1) Multiplying both sides by  1
t = 19
Check:
36  t = 17
36  19 | 17
17 = 17
The solution is 19.
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Slide 2- 13
Combining Like Terms
If like terms appear on the same side of an equation,
we combine them and then solve.
Should like terms appear on both sides of an equation,
we can use the addition principle to rewrite all like
terms on one side.
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Slide 2- 14
Example Solve. 5x + 4x = 36
Solution
5x + 4x = 36
9x = 36
9 x 36

9
9
x=4
Check:
Combining like terms
Dividing both sides by 9
Simplifying
5x + 4x = 36
5(4) + 4(4) | 36
20 + 16 | 36
36 = 36
The solution is 4.
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Slide 2- 15
Example Solve. 4x + 7  6x = 10 + 3x + 12
Solution: 4x + 7  6x = 10 + 3x + 12
2x + 7 = 22 + 3x
2x + 7  7 = 22 + 3x  7
2x = 15 + 3x
2x  3x = 15 + 3x  3x
Combining like terms
Subtracting 7 from both
sides
Simplifying
Subtracting 3x from both
sides
5x = 15
5 x 15

5 5
Dividing both sides by 5
x = 3
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Slide 2- 16
Check:
4x + 7  6x = 10 + 3x + 12
4(3) + 7 6(3) | 10 + 3(3) + 12
12 + 7 + 18 | 10  9 + 12
13 = 13
The solution is 3.
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Slide 2- 17
Example Solve. 4 (6 x  1)  8
5
Solution:
4
(6 x  1)  8
5
5 4
5
 (6 x  1)  8 
4 5
4
6x  1  10
6x 1 1  10 1
6x  9
6x 9

6
6
3
x
2
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Slide 2- 18
An Equation-Solving Procedure
1. Use the multiplication principle to clear any
fractions or decimals. (This is optional, but can ease
computations.
2. If necessary, use the distributive law to remove
parentheses. Then combine like terms on each side.
3. Use the addition principle, as needed, to isolate all
variable terms on one side. Then combine like terms.
4. Multiply or divide to solve for the variable, using
the multiplication principle.
5. Check all possible solutions in the original equation.
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Slide 2- 19
Contradictions and Identities
An identity is an equation that is true for all
replacements that can be used on both sides of the
equation.
A contradiction is an equation that is never true.
A conditional equation is sometimes true and
sometimes false, depending on what the
replacement is.
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Slide 1- 20
Example
a) Solve: 8x  3  2 x  3(2 x  1)
Solution:
8x  3  2 x  3(2 x  1)
8x  3  2 x  6 x  3
8x  3  8x  3
The equation is true regardless of the choice for x, so
all real numbers are solutions. The equation is an
identity.
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Slide 1- 21
Example
b) Solve: x  3  2 x  3( x  5)
Solution:
x  3  2 x  3( x  5)
x  3  2 x  3x  15
x  3  x  15
3  15
The equation is false for any choice of x, so there is no
solution for this equation. The equation is a
contradiction.
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Slide 1- 22
Example
c) Solve: x  11x  3( x  21)
Solution:
x  11x  3( x  21)
x  11x  3x  63
x  8x  63
9 x  63
x7
There is one solution, 7. For all other choices of x, the
equation is false. The equation is a conditional equation.
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Slide 1- 23
2.3
Formulas

Evaluating Formulas

Solving for a Variable
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Many applications of mathematics involve
relationships among two or more quantities. An
equation that represents such a relationship will use
two or more letters and is known as a formula.
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Slide 2- 25
Example
1
M  t
5
The formula
can be used to determine how far
M, in miles, you are from lightening when its thunder
takes t seconds to reach your ears. If it takes 5 seconds
for the sound of thunder to reach you after you have
seen the lightening, how far away is the storm?
Solution We substitute 5 for t and calculate M.
1
M  t
5
1
M  5
5
M 1
The storm is 1 mile away.
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Slide 2- 26
Example Circumference of a circle.
The formula C = d gives the circumference C
of a circle with diameter d. Solve for d.
Solution
C = d
C d

 
C
d

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d
Slide 2- 27
To Solve a Formula for a Given Variable
1. If the variable for which you are solving
appears in a fraction, use the multiplication
principle to clear fractions.
2. Isolate the term(s), with the variable for
which you are solving on one side of the
equation.
3. If two or more terms contain the variable for
which you are solving, factor the variable out.
4. Multiply or divide to solve for the variable in
question.
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Slide 2- 28
Example
Solution
Solve for y: x = wy + zy  6
x = wy + zy  6
x = wy + zy  6
We want this letter alone.
x + 6 = wy + zy
Adding 6 to both sides
x + 6 = y(w + z)
Factoring
x6
y
w z
Dividing both sides by w + z
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Slide 2- 29
2.4
Applications with Percent

Converting Between Percent Notation and
Decimal Notation

Solving Percent Problems
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Percent Notation
n% means
n
1
, or n 
, or n  0.01.
100
100
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Slide 2- 31
Example
Convert to decimal notation: a) 34%
b) 7.6%
Solution
a) 34% = 34  0.01
= 0.34
b) 7.6% = 7.6  0.01
= 0.076
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Slide 2- 32
To convert from percent notation to decimal
notation, move the decimal point two places to
the left and drop the percent symbol.
Example
Convert the percent notation in the following
sentence to decimal notation: Marta received a 75%
on her first algebra test.
Solution
75%  75.0%
0.75.0
75% = 0.75
Move the decimal point two places to the left.
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Slide 2- 33
To convert from decimal notation to percent
notation, move the decimal point two places to
the right and write a percent symbol.
Example
a) 3.24
Convert to percent notation:
3
b) 0.2
c)
8
Solution
a) We first move the decimal point two places to the
right: 3.24. and then write a % symbol: 324%
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Slide 2- 34
Solution continued
b) We first move the decimal point two places to the
right (recall that 0.2 = 0.20): 0.20.
and then write a % symbol: 20%
c) Note that 3/8 = 0.375. We move the decimal point
two places to the right: 0.37.5
and then write a % symbol: 37.5%
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Slide 2- 35
Key Word in Percent Translations
“Of” translates to “•” or “”.
“What” translates to a variable.
“Is” or “Was” translates to “=”.
% translates to
1
"
", or "  0.01".
100
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Slide 2- 36
Example
What is 13% of 72?
Solution
Translate:
What is 13% of 72?
  
a
 
 0.13  72
a  9.36
Thus, 9.36 is 13% of 72. The answer is 9.36.
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Slide 2- 37
Example
9 is 12 percent of what?
Solution
Translate: 9
is 12% of what?
    
9 = 0.12  w
9
w
0.12
75  w
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Slide 2- 38
Example
What percent of 60 is 27?
Solution
Translate: What percent of 60 is 27?
n

60  27
60
n
27
n  0.45  45%
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Slide 2- 39
Example
To complete her water safety course instruction,
Catherine must complete 45 hours of instruction. If she
has completed 75% of her requirement, how many
hours has Catherine completed?
Solution
Rewording: What is 75% of 45?
Translating: a = 0.75  45
a = 33.75
Catherine has completed 33.75 hours of instruction.
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Slide 2- 40
2.5
Problem Solving

Five Steps for Problem Solving

Applying the Five Steps
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Five Steps for Problem Solving in Algebra
1. Familiarize yourself with the problem.
2. Translate to mathematical language. (This
often means writing an equation.)
3. Carry out some mathematical manipulation.
(This often means solving an equation.)
4. Check your possible answer in the original
problem.
5. State the answer clearly, using a complete
English sentence.
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Slide 2- 42
To Become Familiar with a Problem
1. Read the problem carefully. Try to visualize the
problem.
2. Reread the problem, perhaps aloud. Make sure
you understand all important words.
3. List the information given and the question(s) to
be answered. Choose a variable (or variables) to
represent the unknown and specify what the
variable represents. For example, let L = length in
centimeters, d = distance in miles, and so on.
4. Look for similarities between the problem and
other problems you have already solved.
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Slide 2- 43
To Become Familiar with a Problem (continued)
5. Find more information. Look up a formula in a
book, at the library, or online. Consult a reference
librarian or an expert in the field.
6. Make a table that uses all the information you have
available. Look for patterns that may help in the
translation.
7. Make a drawing and label it with known and
unknown information, using specific units if given.
8. Think of a possible answer and check the guess.
Note the manner in which the guess is checked.
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Slide 2- 44
Example
The apartments in Wanda’s apartment house are
consecutively numbered on each floor. The sum of her
number and her next door neighbor’s number is 723.
What are the two numbers?
Solution
1. Familiarize. The apartment numbers are
consecutive integers.
Let x = Wanda’s apartment
Let x + 1 = neighbor’s apartment
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Slide 2- 45
2. Translate.
Rewording: First integer plus second integer is 723
Translating:
3. Carry out.



x

 x 1


 723
x + (x + 1) = 723
2x + 1 = 723
2x = 722
x = 361
If x is 361, then x + 1 is 362.
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Slide 2- 46
4. Check. Our possible answers are 361 and 362.
These are consecutive integers and the sum is
723.
5. State. The apartment numbers are 361 and 362.
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Slide 2- 47
Example
Digicon prints digital photos for $0.12 each plus $3.29
shipping and handling. Your weekly budget for the school
yearbook is $22.00. How many prints can you have made
if you have $22.00?
Solution
1. Familiarize. Suppose the yearbook staff takes 220
digital photos. Then the cost to print them would be
the shipping charge plus $0.12 times 220.
$3.29 + $0.12(220) which is $29.69. Our guess
of 220 is too large, but we have familiarized ourselves
with the way in which the calculation is made.
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Slide 2- 48
2. Translate.
Rewording:
Translating:
3. Carry out.
Shipping
plus

$3.29


photo cost
is


0.12( x ) 
$22

22
3.29  0.12x  22
0.12x  18.71
x  155.9  155
4. Check. Check in the original problem. $3.29 + 155(0.12) =
$21.89, which is less than $22.00.
5. State. The yearbook staff can have 155 photos printed per
week.
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Slide 2- 49
Example
You are constructing a triangular kite. The second
angle of the kite is three times as large as the first. The
third angle is 10 degrees more than the first. Find the
measure of each angle.
Solution
1. Familiarize. Make a drawing and
write in the given information.
3x
x + 10
x
2. Translate. To translate, we need to recall that the
sum of the measures of the angles in a triangle is
180 degrees.
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Slide 2- 50
2. Translate (continued).
Measure of
first angle

x
measure of
measure of
+
second angle
+
third angle
is
180



3x



x  10 



180

3. Carry out.
x  3x   x  10  180
5x  10  180
5x  170
x  34
The measures for the angles appear to be:
first angle:
x = 34
second angle: 3x = 3(34) = 102;
third angle: x + 10 = 34 + 10 = 44
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Slide 2- 51
4. Check. Consider 34, 102 and 44 degrees. The sum of these
numbers is 180 degrees and the second angle is three times
the first angle. The third angle is 10 degrees more than the
first. These numbers check.
5. State. The measures of the angles are 34, 44 and 102
degrees.
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Slide 2- 52
2.6






Solving Inequalities
Solutions of Inequalities
Graphs of Inequalities
Set Builder and Interval Notation
Solving Inequalities Using the Addition
Principle
Solving Inequalities Using the
Multiplication Principle
Using the Principles Together
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Solutions of Inequalities
An inequality is a number sentence containing > (is
greater than), < (is less than),  (is greater than or
equal to), or  (is less than or equal to).
Example
Determine whether the given number is a
solution of x < 5: a) 4 b) 6
Solution
-4
-3
-2
-1
0
1
2
3
4
5
6
a) Since 4 < 5 is true, 4 is a solution.
b) Since 6 < 5 is false, 6 is not a solution.
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Slide 2- 54
Solutions of Inequalities
An inequality is any sentence containing
, , ,  , or  .
Examples:
3x  2  7, c  7, and 4x  6  3.
Any value for a variable that makes an inequality true is
called a solution. The set of all solutions is called the
solution set. When all solutions of an inequality are
found, we say that we have solved the inequality.
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Slide 4- 55
Example
Determine whether 5 is a solution to
3x  2  7.
Solution
We substitute to get 3(5) + 2 > 7, or 17 >7, a
true statement. Thus, 5 is a solution.
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Slide 4- 56
The graph of an inequality is a visual representation of the
inequality’s solution set. An inequality in one variable can be
graphed on a number line.
Example
Graph x < 2 on a number line.
Solution
)
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
Note that in set-builder notation the solution is
x | x  2.
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Slide 4- 57
Interval Notation
We can write solutions of an inequality in one variable
using interval notation. Interval notation uses
parentheses, ( ), and brackets, [ ].
If a and b are real numbers such that a < b, we define the open
interval (a, b) as the set of all numbers x for which a < x < b.
Thus,
(a, b)   x | a  x  b.
(a, b)
(
a
)
b
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Slide 4- 58
Interval Notation
The closed interval [a, b] is defined as the set of all numbers
x for which
a  x  b.
Thus,
[a, b]   x | a  x  b.
[a, b]
[
a
]
b
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Slide 4- 59
Interval Notation
There are two types of half-open intervals, defined as
follows:
1. (a, b]   x | a  x  b.
(a, b]
(
a
]
b
2. [a, b)   x | a  x  b.
[a, b)
[
a
)
b
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Slide 4- 60
Interval Notation
We use the symbols
represent positive and
 and to

negative infinity, respectively. Thus the notation (a, )
represents the set of all real numbers greater than a, and (

a) represents the set of all numbers less than a.
,

(, a)
(
a
(a, )
)
a
The notations (– , 
a] and [a,
include the endpoint a.
) are
used when we want to
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Slide 4- 61
The Addition Principle for
Inequalities
For any real numbers a, b, and c:
a < b is equivalent to a + c < b + c;
a  b is equivalent to a + c  b + c;
a > b is equivalent to a + c > b + c;
a  b is equivalent to a + c  b + c.
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Slide 2- 62
Example
Solve and graph x – 2 > 7.
Solution
x–2>7
x–2+2>7+2
x>9
x | x  9, or (9, ).
The solution set is
(
5
6
7
8
9
10 11 12 13 14 15 16 17
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Slide 4- 63
Example
Solve 4x  1  x  10 and then graph the solution.
Solution
4x  1  x  10
4x  1 + 1  x  10 + 1 Adding 1 to both sides
Simplifying
4x  x  9
Subtracting x from both sides
4x  x  x  x  9
Simplifying
3x  9
Dividing both sides by 3
x  3
The solution set is {x|x  3}.
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 2- 64
The Multiplication Principle for
Inequalities
For any real numbers a and b, and for any positive
number c:
a < b is equivalent to ac < bc, and
a > b is equivalent to ac > bc.
For any real numbers a and b, and for any negative
number c:
a < b is equivalent to ac > bc, and
a > b is equivalent to ac < bc.
Similar statements hold for  and .
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Slide 2- 65
Example
Solve and graph each inequality:
a) 1 x  4
b) 4y < 20
7
Solution
a) 1 x  4
7
1
4 x  74
3
Multiplying both sides by 4
x  28
Simplifying
The solution set is {x|x  28}. The graph is shown below.
5
10
15
20
25
]
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30
Slide 2- 66
b) 4y < 20
4 y 20

4 4
Dividing both sides by 4
At this step, we reverse the
inequality, because 4 is negative.
y  5
The solution set is {y|y > 5}. The graph is shown
below.
-6
(
-5
-4
-3
-2
-1
0
1
2
3
4
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Slide 2- 67
Example
Solve. 3x  3 > x + 7
Solution
3x  3 > x + 7
3x  3 + 3 > x + 7 + 3
3x > x + 10
3x  x > x  x + 10
2x > 10
Adding 3 to both sides
Simplifying
Subtracting x from both sides
Simplifying
2 x 10

2
2
Dividing both sides by 2
Simplifying
x>5
The solution set is {x|x > 5}.
-2
-1
0
1
2
3
4
(
5
6
7
8
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Slide 2- 68
Example
Solution
Solve. 15.4  3.2x < 6.76
15.4  3.2x < 6.76
100(15.4  3.2x) < 100(6.76)
100(15.4)  100(3.2x) < 100(6.76)
1540  320x < 676
320x < 676 1540
320x < 2216
x
2216
320
x > 6.925
Remember to reverse
the symbol!
The solution set is {x|x > 6.925}.
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 2- 69
Example
Solve: 5(x  3)  7x  4(x  3) + 9
Solution
5(x  3)  7x  4(x  3) + 9
5x  15  7x  4x  12 + 9
2x  15  4x  3
2x  15 + 3  4x  3 + 3
2x  12  4x
2x + 2x  12  4x + 2x
12  6x
2  x
The solution set is {x|x  2}.
-8
-7
-6
-5
-4
-3
]
-2
-1
Using the distributive law to
remove parentheses
Simplifying
Adding 3 to both sides
Adding 2x to both sides
Dividing both sides by 6
0
1
2
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Slide 2- 70
2.7
Solving Applications with
Inequalities

Translating to Inequalities

Solving Problems
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Important Words
Sample Sentence
Translation
b  16
is at least
Brian is at least 16 years old
is at most
At most 3 students failed the course
s3
cannot exceed
To qualify, earnings cannot exceed
$5000
e  $5000
must exceed
The speed must exceed 20 mph
s > 20
is less than
Nicholas is less than 60 lb.
n < 60
is more than
Chicago is more than 300 miles
away.
c > 300
is between
The movie is between 70 and 120
minutes.
70 < m < 120
no more than
The calf weighs no more than 560
lb.
w  560
no less than
Carmon scored no less than 9.4.
c  9.4
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 2- 72
Translating “at least” and “at most”
The quantity x is at least some amount q: x  q.
(If x is at least q, it cannot be less than q.)
The quantity x is at most some amount q: x  q.
(If x is at most q, it cannot be more than q.)
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 2- 73
Example
Lazer Line charges $65 plus $45 per hour for copier repair.
Jonas remembers being billed less than $150. How many hours
was Jonas’ copier worked on?
Solution
1. Familiarize. Suppose the copier was worked on for
4 hours. The cost would be $65 + 4($45), or $245.
This shows that the copier was worked on for less
than 4 hours. Let h = the number of hours.
2. Translate.
Rewording: Initial fee plus hours less than
Translating:

65



45h


Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
150

150
Slide 2- 74
3. Carry out.
65 + 45h  150
45h  85
85
h
45
8
h 1
9
4. Check. Since the time represents hours, we round down to
one hour. If the copier was worked on for one hour, the cost
would be $110, and if worked on for two hours the cost
would exceed $150.
5. State. Jonas’ copier was worked on for less than two hours.
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 2- 75
Average, or Mean
To find the average or mean of a set of
numbers, add the numbers and then divide by
the number of addends.
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Slide 2- 76
Example
Samantha has test grades of 86, 88, and 78 on her first three
math tests. If she wants an average of at least 80 after the fourth
test, what possible scores can she earn on the fourth test?
Solution
1. Familiarize. Suppose she earned an 85 on her fourth test.
Her test average would be
86  88  78  85
 84.25
4
This shows she could score an 85. Let’s let x represent the
fourth test score.
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Slide 2- 77
2. Translate.
Rewording:
should be
Average test scores at least


Translating:
3. Carry out.
86  88  78  x
4

80

80
86  88  78  x
 80
4
86  88  78  x
4
 80  4
4
252  x  320
x  68
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Slide 2- 78
4. Check.
As a partial check, we show that Samantha can earn a 68 on |
the fourth test and average 80 for the four tests.
86  88  78  68 320

 80.
4
4
5. State.
Samantha’s test average will not drop below 80 if she earns
at least a 68 on the fourth test.
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Slide 2- 79