Transcript Section 2.4

2.4 An Introduction
to Applications of
Linear Equations
Objective 1
Learn the six steps for solving applied
problems.
Slide 2.4-3
Learn the six steps for solving 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.
Solving an Applied Problem
Step 1: Read the problem carefully. What information is given? What are
you asked to find?
Step 2: Assign a variable to represent the unknown value. Use a sketch,
diagram, or table, as needed. 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. Label it appropriately. Does it seem reasonable?
Step 6: Check the answer in the words of the original problem.
Slide 2.4-4
Objective 2
Solve problems involving unknown
numbers.
Slide 2.4-5
Learn the six steps for solving applied problems. (cont’d)
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.
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.
Slide 2.4-6
CLASSROOM
EXAMPLE 1
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.
Equation:
Solve:
Let x = the number.
9x  5  x  x 19  x
8x  5  5  19  5
8 x 24

8
8
x  3
9x  5  x 19
The number is −3.
When solving an equation, use solution set notation to write the answer. When
solving an application, state the answer in a sentence.
Slide 2.4-7
Objective 3
Solve problems involving sums of
quantities.
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.
PROBLEM-SOLVING HINT
To solve problems involving sums of quantities, choose a variable to
represent one of the unknowns. Then represent the other quantity in
terms of the same variable, using information from the problem.
Slide 2.4-9
CLASSROOM
EXAMPLE 2
Finding the Numbers of Olympic Medals
In the 2006 Winter Olympics in Torino, Italy, Canada won 5 more medals
than Norway. The two countries won a total of 43 medals. How many
medals did each country win? (Source: U.S. Olympic Committee.)
Solution:
Let x = the number of medals Norway won.
Let x + 5 = the number of medals Canada won.
43  x   x  5
43  5  2x  5  5
38 2x

2
2
x  19
19  5  24
Norway won 19 medals and Canada won 24 medals.
Slide 2.4-10
Solve problems involving sums of quantities. (cont’d)
The problem in Example 2 could also be solved by letting x represent the
number of medals Canada won. Then x − 5 would represent the number of
medals Norway won. The equation would be
43  x   x  5
The solution to this equation is 24, which is the number of medals Canada
won. The number of Norwegian medals would be 24 − 5 = 19. The answers
are the same, whichever approach is used, even though the equation and
its solution are different.
The nature of the applied problem restricts the set of possible solutions. For
1
example, an answer such as −33 medals or 25 medals should
be recognized as
2
inappropriate.
Slide 2.4-11
CLASSROOM
EXAMPLE 3
Finding the Number of Orders for Tea
On that same day, the owner of Terry’s Coffeehouse found that the number
of orders for croissants was 1 the number of muffins. If the total number
6
for the two breakfast rolls was 56, how many orders were placed for
croissants?
Solution:
Let x = the number of muffins.
Then
1
x = the number of croissants.
6
1
56  x  x
6 1
56  6   x  6   x  6 
6
8 croissants were ordered.
336  6x  x
336 7 x

7
7
x  48
1
 48   8
6
Slide 2.4-12
Solve problems involving sums of quantities. (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
CLASSROOM
EXAMPLE 4
Analyzing a Gasoline-Oil Mixture
At a meeting of the local computer user group, each member brought two
nonmembers. If a total of 27 people attended, how many were members
and how many were nonmembers?
Solution:
Let x = number of members.
Then 2x = number of nonmembers.
2x  x  27
3x 27

3
3
x 9
2  9  18
There were 9 members and 18 nonmembers at the meeting.
Slide 2.4-14
Solve problems involving sums of quantities. (cont’d)
PROBLEM SOLVING HINT
Sometimes it is necessary to find three unknown quantities. When the three
unknowns are compared in pairs, let the variable represent the unknown
found in both pairs.
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CLASSROOM
EXAMPLE 5
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.
Solution:
Let x = the length of the middle-sized piece,
then x +10 = the longest piece,
and x − 5 = the shortest piece.
x   x  10   x  5  50
3x  5  5  50  5
3 x 45

3
3
x  15
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.
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Objective 4
Solve problems involving consecutive
integers.
Slide 2.4-17
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 greater
consecutive integer.
Consecutive even integers, such as 8 and 10, differ by 2.
Consecutive odd integers, such as 9 and 11, also differ by 2.
In general if x represents an even integer, x + 2 represents the greater
consecutive integer. The same holds true for odd integers; that is if x is an odd
integer, x + 2 is the greater odd integer.
Slide 2.4-18
Solve problems involving consecutive integers. (cont’d)
PROBLEM SOLVING HINT
If x = the lesser integer, then, for any
two consecutive integers, use
x, x  1;
two consecutive even integers, use
x, x  2;
two consecutive odd integers, use
x,x  2.
Slide 2.4-19
CLASSROOM
EXAMPLE 6
Finding Consecutive Integers
Two pages that face each other have 569 as the sum of their page
numbers. What are the page numbers?
Solution:
Let x = the lesser page number.
Then x + 1= the greater page number.
x   x  1  569
2x 1 1  569 1
2 x 568

2
2
x  284
284  1  285
The lesser page number is 284, and the greater page number is 285.
It is a good idea to use parentheses around x + 1, (even though they are not
necessary here).
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CLASSROOM
EXAMPLE 7
Finding Consecutive Odd Integers
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.
6 x   x  2  86
7x  2  2  86  2
7 x 84

7
7
x  12
12  2  14
The lesser integer is 12 and the greater integer is 14.
Slide 2.4-21
Objective 5
Solve problems involving
supplementary and complementary
angles.
Slide 2.4-22
Solve problems involving supplementary and
complementary angles.
An angle can be measured by a unit called the degree (°), which is 1 of a
360
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 2.4-23
Solve problems involving supplementary and complementary
angles. (cont’d)
PROBLEM-SOLVING HINT
If x represents the degree measure of an angle, then
90 − x represents the degree measure of its complement,
180 − x represents the degree measure of is supplement.
Slide 2.4-24
CLASSROOM
EXAMPLE 8
Finding the Measure of an Angle
Find the measure of an angle whose complement is eight times its measure.
Solution:
Let x = the degree measure of the angle.
Then 90 − x = the degree measure of its complement.
90  8x  x
90  9x
90 9x

9
9
x  10
The measure of the angle is 10°.
Slide 2.4-25
CLASSROOM
EXAMPLE 9
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°.
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  x   180  x   174
270  2x  270  174  270
2 x 96

2
2
x  48
The measure of the angle is 48°.
Slide 2.4-26