2.4 An Introduction to Applications of Linear Equations

Objective 1

Learn the six steps for solving applied problems.

Slide 2.4-3

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 ProblemStep 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

Learn the six steps for solving applied problems.

Objective 2

Solve problems involving unknown numbers.

Slide 2.4-5

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

Learn the six steps for solving applied problems. (cont’d)

When solving an equation, use solution set notation to write the answer. When solving an application, state the answer in a sentence.

Equation:

Let x = the number.

CLASSROOM EXAMPLE 1

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 19x x 9 5 19x xx x 8 5 195 5x

8 2

8 8

4x

3x The number is −3.

Slide 2.4-7

Finding the Value of an Unknown Number

Solve:

Objective 3

Solve problems involving sums of quantities.

Slide 2.4-8

A common type of problem in elementary algebra involves finding two quantities when the sum of the quantities is known.

PROBLEM-SOLVING HINTTo 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

Solve problems involving sums of quantities.

Solution:Let x = the number of medals Norway won.

CLASSROOM EXAMPLE 2

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.)

43 5x x 43 25 55x 38

2 2

2x

19x

Norway won 19 medals and Canada won 24 medals.

Let x + 5 = the number of medals Canada won.

19 5 24

Slide 2.4-10

Finding the Numbers of Olympic Medals

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 5x x

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 example, an answer such as −33 medals or 25 medals should be recognized as inappropriate.

1

2

Slide 2.4-11

Solve problems involving sums of quantities. (cont’d)

CLASSROOM EXAMPLE 3

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

156

6x x

6 6 61

566

x x

336 6x x 336 7

7 7

x

48x

1

648 8

8 croissants were ordered.

1

6

Slide 2.4-12

Finding the Number of Orders for Tea

PROBLEM SOLVING HINTIn 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

Solve problems involving sums of quantities. (cont’d)

Solution:Let x = number of members.Then 2x = number of nonmembers.

3 27

3 3

x

2 27x x

9x

2 9 18

CLASSROOM EXAMPLE 4

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?

There were 9 members and 18 nonmembers at the meeting.

Slide 2.4-14

Analyzing a Gasoline-Oil Mixture

PROBLEM SOLVING HINTSometimes 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.

Slide 2.4-15

Solve problems involving sums of quantities. (cont’d)

Solution:Let x = the length of the middle-sized piece, then x +10 = the longest piece,and x − 5 = the shortest piece.

10 5 50x x x 55 50 53x 3 5

3 3

4x

CLASSROOM EXAMPLE 5

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.

15x

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.

Slide 2.4-16

Dividing a Pipe into Pieces

Objective 4

Solve problems involving consecutive integers.

Slide 2.4-17

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.

PROBLEM SOLVING HINTIf x = the lesser integer, then, for any

two consecutive integers, use

two consecutive even integers, use

two consecutive odd integers, use

, 1;x x

, 2;x x

, 2.x x

Slide 2.4-19

Solve problems involving consecutive integers. (cont’d)

Solution:Let x = the lesser page number.Then x + 1= the greater page number.

1 569x x 1 12 1 569x 2 8

2 2

56x

CLASSROOM EXAMPLE 6

Two pages that face each other have 569 as the sum of their page numbers. What are the page numbers?

284x

The lesser page number is 284, and the greater page number is 285.

184 82 2 5

It is a good idea to use parentheses around x + 1, (even though they are not necessary here).

Slide 2.4-20

Finding Consecutive Integers

6 2 86x x

CLASSROOM EXAMPLE 7

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

8x

22 86 27x

12 2 14

12x

Slide 2.4-21

Finding Consecutive Odd Integers

Objective 5

Solve problems involving supplementary and complementary angles.

Slide 2.4-22

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°.

1

360

Slide 2.4-23

Solve problems involving supplementary and complementary angles.

PROBLEM-SOLVING HINTIf 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

Solve problems involving supplementary and complementary angles. (cont’d)

Solution:Let x = the degree measure of the angle.Then 90 − x = the degree measure of its complement.

90 8x x 90 9x90

9 9

9x

CLASSROOM EXAMPLE 8Find the measure of an angle whose complement is eight times its measure.

10x

The measure of the angle is 10°.Slide 2.4-25

Finding the Measure of an Angle

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 174x x 2702 270 2 70174x 2 96

2 2

x

CLASSROOM EXAMPLE 9

Find the measure of an angle such that the sum of the measures of its complement and its supplement is 174°.

48x

The measure of the angle is 48°.Slide 2.4-26

Finding the Measure of an Angle