mba10_ppt_0204 (1).ppt

Chapter 2
Section 4
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

An Introduction to Applications of
Linear Equations
Learn the six steps for solving applied
problems.
Solve problems involving unknown numbers.
Solve problems involving sums of quantities.
Solve problems involving supplementary and
complementary angles.
Solve problems involving consecutive integers.
1
4
3
2
5
2.4

Objective 1
Slide 2.4 - 3
Learn the six steps for solving
applied problems.

Learn the six steps for solving applied problems.
We now look at how algebra is used to solve applied problems.
While there is no one specific method that enables you to solve all
kinds of applied problems, the following six-step method is often
applicable.
Step 1: Read the problem carefully until you understand what
is given and what is to be found.
Step 2: Assign a variable to represent the unknown value, using
diagrams or tables as needed. Write down what the variable
represents. If necessary, express any other unknown values
in terms of the variable.
Step 3: Write an equation using the variable expression(s).
Step 4: Solve the equation.
Step 5: State the answer. Does it seem reasonable?
Step 6: Check the answer in the words of the original problem.
Slide 2.4 - 4

The third step in solving an applied problem is often the
hardest. To translate the problem into an equation, write the
given phrases as mathematical expressions. Replace any words
that mean equals or same with an = sign.
Slide 2.4 - 5
Other forms of the verb “to be,” such as is, are, was, and were,
also translate this way. The = sign leads to an equation to be
solved.
Learn the six steps for solving applied
problems. (cont’d)

Objective 2
Slide 2.4 - 6
Solve problems involving
unknown numbers.

EXAMPLE 1
Slide 2.4 - 7
When solving an equation, use solution set notation to write the answer.
When solving an application, state the answer in a sentence.
Solution:
Let x = the number.
Finding the Value of an
Unknown Number
If 5 is added to the product of 9 and a number, the
result is 19 less than the number. Find the number.
9 5 19
x x
  
9 5 19
x x
x x
  
 
8 5 19
5 5
x  
 
8 2
8 8
4
x 

3
x  
The number is −3.

Objective 3
Slide 2.4 - 8
Solve problems involving sums of
quantities.

A common type of problem in elementary algebra involves
finding two quantities when the sum of the quantities is known.
In Example 9 of Section 2.3, we prepared for this type of
problem, by writing mathematical expressions for two related
unknown quantities.
Solve problems involving sums of quantities.
Slide 2.4 - 9
PROBLEM-SOLVING HINT
To solve problems involving sums of quantities, choose a
variable to represent one of the unknowns. Represent the other
quantity in terms of the same variable, using information from
the problem. Write an equation based on the words of the
problem.

EXAMPLE 2
Slide 2.4 - 10
Slide 2.4 - 10
Solution:
Let x = the number of medals Norway won.
Finding Numbers of Olympics
Medals
In the 2006 Winter Olympics in Torino, Italy, the United States
won 6 more medals than Norway. The two countries won a
total of 44 medals. How many medals did each country win?
(Source: U.S. Olympic Committee.)
 
44 6
x x
  
44 2
6 6
6
x

  
38
2 2
2x

19
x 
Norway won 19 medals and the
U.S. won 25 medals.
Let x + 6 = the number of medals the U.S. won.
19 6 25
 

Solve problems involving sums of quantities. (cont’d)
Slide 2.4 - 11
The problem in example 2 could also be solved by letting x
represent the number of medals the United States won. Then
x − 6 would represent the number of medals Norway won. The
equation would be
 
44 6
x x
  
The solution to this equation is 25, which is the number of
medals the U.S. won. The number of Norwegian medals would
be 25 − 6 = 19. The answers are the same, whichever approach
is used.
The nature of the applied problem restricts the set of possible solutions.
For example, an answer such as −33 medals or 25 medals should be
recognized as inappropriate.
1
2

EXAMPLE 3
Slide 2.4 - 12
Finding the Number of Orders
for Croissants
On that same day, the owner of Terry’s Coffeehouse
found that the number of orders for croissants was
the number of muffins. If the total number for the two
breakfast rolls was 56, how many orders were placed
for croissants?
Solution:
Let x = the number of muffins.
Then x = the number of croissants.
1
6
1
56
6
x x
 
     
6 6 6
1
56
6
x x
 
336 6x x
 
336 7
7 7
x

48
x 
 
1
6
48 8

8 croissants were ordered.
1
6

Finding the Number of Orders for
Croissants (cont’d)
PROBLEM SOLVING HINT
In Example 3, it was easier to let the variable represent the
quantity that was not specified. This required extra work in Step
5 to find the number of orders for croissants. In some cases, this
approach is easier than letting the variable represent the quantity
that we are asked to find.
Slide 2.4 - 13

EXAMPLE 4
Solution:
Let x = number of members.
Then 2x = number of nonmembers.
Slide 2.4 - 14
3 27
3 3
x

2 27
x x
 
9
x 
 
2 9 18

Analyzing the Mixture of a
Computer User Group
At a meeting of the local computer user group, each
member brought two nonmembers. If a total of 27
people attended, how many were nonmembers?
There were 9 members and 18
nonmembers at the meeting.

Dividing a Board into Pieces
Sometimes it is necessary to find three unknown quantities in
an applied problem. Frequently, the three unknowns are
compared in pairs. When this happens it is usually easiest to let
the variable represent the unknown found in both pairs.
Slide 2.4 - 15

EXAMPLE 5
Slide 2.4 - 16
Solution:
Let x = the length of the middle-sized piece,
then x +10 = longest piece,
and x − 5 = shortest piece.
   
10 5 50
x x x
    
5
5 50 5
3x 
 
3 5
3 3
4
x

Dividing a Pipe into Pieces
A piece of pipe is 50 in. long. It is cut into three pieces.
The longest piece is 10 in. more than the middle-sized
piece, and the shortest piece measures 5 in. less than the
middle-sized piece. Find the lengths of the three pieces.
15
x 
15 10 25
 
15 5 10
 
The shortest piece is 10 in.,
the middle-size piece is 15
in., and the longest is 25 in.

Objective 4
Slide 2.4 - 17
supplementary and complementary
angles.

complementary angles.
An angle can be measured by a unit called the degree (°),
which is of a complete rotation.
Two angles whose sum is 90° are said to be complementary,
or complements of each other. An angle that measures 90° is a
right angle.
Two angles who sum is 180° are said to be supplementary, or
supplements of each other. One angle supplements the other to
form a straight angle of 180°.
Slide 2.4 - 18
1
360

complementary angles. (cont’d)
If x represents the degree measure of an angle, then
90 − x represents the degree measure of its complement,
and 180 − x represents the degree measure of is supplement.
Slide 2.4 - 19

EXAMPLE 6
Slide 2.4 - 20
Solution:
Let x = the degree measure of the angle.
Then 90 − x = the degree measure of its complement,
and 180 − x = the degree measure of its supplement.
   
90 180 174
x x
   
270
2 2
70 2 70
174
x
 
 
2 96
2 2
x
 

 
Finding the Measure of an Angle
Find the measure of an angle such that the sum of the
measures of its complement and its supplement is 174°.
48
x 
The measure of the angle is 48°.

Objective 5
Slide 2.4 - 21
consecutive integers.

Solve problems involving consecutive integers.
Two integers that differ by 1 are called consecutive integers.
For example, 3 and 4, 6 and 7, and −2 and −1 are pairs of
consecutive integers. In general, if x represents an integer, x
+ 1 represents the next larger consecutive integer.
Slide 2.4 - 22
Consecutive even integers, such as 8 and 10, differ by 2.
Similarly, consecutive odd integers, such as 9 and 11, also differ
by 2. In general if x represents an even integer, x + 2 represents
the larger consecutive integer. The same holds true for odd
integers; that is if x is an odd integer, x + 2 is the larger odd
integer.

Solve problems involving consecutive
integers. (cont’d)
In solving consecutive integer problems, if x = the first integer,
then, for any
two consecutive integers, use
two consecutive even integers, use
two consecutive odd integers, use
Slide 2.4 - 23
, 1;
x x
 
, 2;
x x
 
, 2.
x x
 

EXAMPLE 7
Slide 2.4 - 24
Solution:
Let x = the lesser page number.
Then x + 1= the greater page number.
 
1 569
x x
  
1 1
2 1 569
x 
 
2 8
2 2
56
x

Finding Consecutive Integers
Two back-to-back page numbers in this book have a
sum of 569. What are the page numbers?
284
x 
The lesser page number is 284,
and the greater page number is
285.
1
84 8
2 2 5
 
It is a good idea to use parentheses around x + 1, (even though they are
not necessary here).

 
6 2 86
x x
  
EXAMPLE 8
Slide 2.4 - 25
Find two consecutive even integers such that six times
the lesser added to the greater gives a sum of 86.
Solution:
Let x = the lesser integer.
Then x + 2 = the greater integer.
The lesser integer is 12 and
the greater integer is 14.
7 4
7 7
8
x

2
2 86 2
7x 
 
12 2 14
 
12
x 
Finding Consecutive Even
Integers

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