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Bittinger
Pre-Algebra, 6th edition
Section 5.8 – Applications and Problem Solving
Tom Atwater
Draft 1
Example from Text:
Exercises from Text: 1, 5, 9, 15, 23, 25, 39, 53
Student: No
Video
Audio
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Section 5.8
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Applications and Problem Solving
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INTRODUCTION:
Hi, my name is Tom. Today we are going
to learn how to solve application problems
involving decimals.
READ Objectives.
Objectives
Translate key phrases to algebraic
expressions.
Solve applied problems involving
decimals.
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SAY: To translate problems to equations,
we need to be able to translate phrases to
algebraic expressions. Certain key words
in phrases help direct the translation.
READ Key Words
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Bittinger
Pre-Algebra, 6th edition
Section 5.8 – Applications and Problem Solving
Tom Atwater
Draft 1
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SAY: It is helpful to choose a descriptive
variable to represent the unknown. For
example, w suggests weight and g
suggests the number of gallons of
gasoline.
The following tips are helpful in
translating phrases to algebraic
expressions.
READ Tips
TIPS FOR TRANSLATING PHRASES
TO ALGEBRAIC EXPRESSIONS
•
Use a specific number in the
statement before translating using the
variable.
•
Write down what each variable
represents.
•
Check the translation with another
number to see if it matches the phrase.
•
Be especially careful with order
when subtracting and dividing.
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SAY: Let’s work an example.
READ Exercise 1, 5, 9, 15, 23
Translate to an algebraic expression.
Choice of variables used may vary.
(a)
(b)
(c)
(d)
(e)
Five more than Ron’s age
9 less than c
8 times Nate’s speed
20 less than 4 times a number
5 times the difference of two numbers
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Bittinger
Pre-Algebra, 6th edition
Section 5.8 – Applications and Problem Solving
Tom Atwater
Draft 1
MATH STEPS:
a) Five more than Ron’s age
Let a = Ron’s age
a + 5, or 5 + a
(b) 9 less than c
c–9
(c) 8 times Nate’s speed
Let s = Nate’s speed
8s
(d) 20 less than 4 times a number
let x = the number
4x – 20
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(e) 5 times the difference of two numbers
let x and y = the numbers
5(x – y)
SAY: Next we consider solving applied
problems. We will use the Five Steps for
Problem Solving learned earlier in the
text.
SAY: Let’s work an example.
READ Exercise 25
The movie Avatar took in $2.63 billion in
its lifetime. This is $0.78 billion more
than Titanic took in. How much did
Titanic take in during its lifetime?
MATH STEPS:
1. Familiarize
We let t = how much Titanic took in,
which is less than Avatar
2. Translate.
The problem can be translated to an
equation as follows:
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Bittinger
Pre-Algebra, 6th edition
Section 5.8 – Applications and Problem Solving
Tom Atwater
Draft 1
t + 0.78 = 2.63
3. Solve.
To solve the equation, we subtract 0.78
from both sides.
t + 0.78 – 0.78 = 2.63 – 0.78
t = 1.85
4. Check.
We check by adding.
1.85 + 0.78 = 2.63
5. State
Titanic took in $1.85 billion.
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SAY: Here’s an example that involves
more than one mathematical operation.
READ Exercise 39
Peggy filled her van’s gas tank and noted
that the odometer read 26,342.8. After the
next filling, the odometer read 26,736.7. It
took 19.5 gal to fill the tank. How many
miles per gallon did the van get?
MATH STEPS:
1. Familiarize
First make a drawing
26,342.8 n miles, 19.5 gallons 26,736.7
This is a two-step problem. First, we find
the number of miles that have been driven
between fillups. We let n = the number of
miles driven and m = the number of miles
per gallon.
2. Translate.
The problem can be translated to an
equation as follows:
First Reading plus miles driven is second
reading
26,342.8 + n = 26,736.7
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Bittinger
Pre-Algebra, 6th edition
Section 5.8 – Applications and Problem Solving
Tom Atwater
Draft 1
3. Solve.
To solve the equation, we subtract
26,342.8 from both sides.
26,342.8 + n – 26,342.8 = 26,736.7 –
26,342.8
n = 393.9
Next, we divide the total number of miles
driven by the number of gallons. This
gives us m. The division that corresponds
to the situation is 393.9 ÷ 19.5
19.5 393.9
20.2
195 3939.0
390
390
4. Check.
To check, we first multiply the number of
miles per gallon times the number of
gallons to find the number of miles
driven:
20.2 x 19.5 = 393.9
Then we add 393.9 to 26,342.8 to find the
new odometer reading:
26,342.8 + 393.9 = 26,736.7
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5. State
Peggy got 20.2 miles per gallon.
SAY: This example involves a formula
giving the area of a circle.
READ Area of a Circle
In any circle, a diameter is a segment that
passes through the center of the circle
with endpoints on the circle. A radius is a
segment with one endpoint on the center
and the other endpoint on the circle. The
area, A, of a circle with radius of length r
is given by
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Bittinger
Pre-Algebra, 6th edition
Section 5.8 – Applications and Problem Solving
Tom Atwater
Draft 1
A = π • r2 ,
where π ≈ 3.14.
The length r of a radius of a circle is half
the length d of a diameter.
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r d.
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READ Exercise 53
A round, 6-ft-wide hot tub is being built
into a 12-ft by 30-ft rectangular deck.
How much decking is needed for the
surface of the deck?
MATH STEPS:
1. Familiarize
First make a drawing
Let d = the amount of decking needed
The area of a rectangle is l•w
The area of a circle is A = π • r2 , where π
≈ 3.14
The radius of a circle is half its diameter: r
=½d
The length is 30, the width is 12.
The radius is r = ½ • 6 = 3
2. Translate.
To find the amount of decking needed we
subtract the area of the circle form the
area of the rectangle.
d = l•w – π • r2
d = 30 • 12 – π • 3^2
3. Solve.
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Bittinger
Pre-Algebra, 6th edition
Section 5.8 – Applications and Problem Solving
Tom Atwater
Draft 1
To solve the equation, we carry out the
operations.
d = 360 – π • 3^2
d = 360 – 28.26
d = 331.74
4. Check.
To check, we can repeat our calculations.
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5. State
The amount of material needed for the
decking is 331.74 square feet.
CONCLUSION:
Today we learned how to solve applied
problems involving decimal notation. Go
back and review this section before
beginning the exercise set. If you're
having any trouble with the material see
your instructor immediately. Good Luck.
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