Copyright © Cengage Learning. All rights reserved. Systems of Linear Equations SECTION 2.4.

Copyright © Cengage Learning. All rights reserved.

Systems of Linear Equations

SECTION 2.4

2

Learning Objectives

1 Determine the solution to a system of equationsalgebraically, graphically, and using technology and interpret the real-world meaning of the results

2 Use the substitution and elimination methods to solve linear systems that model real-world scenarios

3 Determine if systems of linear equations are dependent or inconsistent and explain the real-world meaning of the results

3


4


A system of equations is a group of two or more equations. To solve a system of equations means to find values for the variables that satisfy all of the equations in the system.

Systems of equations can involve any number of equations and variables; however, we will limit ourselves to situations containing two variables in this section.

© 2010 Pearson Education, Inc. All rights reserved. Section 8.1, Slide 5


A solution of a pair of linear equations is an ordered pair of numbers that satisfies both equations. The ordered pair (5, 3) is a solution of the linear equations below.

6

Although (7, 2) makes the first equation true in the system…

(2)(7) + (4)(2) = 22

…it does not make the second equation true.

7 – (6)(2) 5

Therefore, it is not a solution for the system.

© 2010 Pearson Education, Inc. All rights reserved. Section 8.1, Slide 6


7© 2010 Pearson Education, Inc. All rights reserved. Section 8.1, Slide 7

Systems of Linear EquationsGraphing a system of two linear equations in two unknowns gives one of three possible situations:

Case 1: Lines intersecting in a single point. The ordered pair that represents this point is the unique solution for the system.



Case 2: Lines that are distinct parallel lines and therefore don’t intersect at all. Because the lines have no common points, this means that the system has no solutions.



Case 3: Two lines that are the same line. The lines have an infinite number of points in common, so the system will have an infinite number of solutions.


Solving Systems Using the Elimination Method• Example: Solve the system.

(continued on next slide)


Solving Systems Using the Elimination Method• Example: Solve the system.

• Solution:Multiply the top equation by 3 and the bottom equation by 2 to get opposite coefficients for y.



Solving Systems Using the Elimination Method

Next add corresponding sides of both equations and the y drops out.


Solve for x.



To find y, substitute 2 for x in either equation of the original system.

Thus, the solution for this system is (2, 1) (Case 1).



• Example: Solve the system




• Example: Solve the system

• Solution:Multiply both sides of the top equation by 4 to clear the fractions.




Multiply the bottom equation by 2 and add the equations to eliminate x from the system:

There are no points common to both lines (Case 2).

A system that has no solutions is said to be inconsistent.


Solving Systems Using the Elimination Method• Example: Solve the system



Solving Systems Using the Elimination Method• Example: Solve the system

•Solution:Multiply the top equation by 10 and the bottom equation by 100 to get rid of the decimals.



Solving Systems Using the Elimination MethodMultiply the top equation by –5, and add the two equations to eliminate x from the system.

The two lines must be the same (Case 3).

A system that has an infinite number of solutions is said to be dependent.

21

Solving a System Using the Elimination Method

The method of eliminating one of the variables to get an equation, which is easier to solve is known as elimination method.

22

Solving a System Using the Substitution Method

23


24

Example 1 – Exploring a System of Equations Using a Table of Values and Substitution.

Suppose we hired a taxi in New York City and in Baltimore.

Table 2.12

25


(a) Write a system of equations to represent the total costs of hiring a taxi in New York City and Baltimore as functions of the number of additional miles traveled beyond the flag drop. (Let m represent the number of miles traveled beyond the flag drop. Let N(m) represent the total costs of hiring a taxi in New York and B(m) represent the total costs of hiring a taxi in Baltimore.) Assume each encounters 5 minutes of wait time during their rides.

26


Solution:

N(m) = 2.00m + 4.50 B(m) = 2.20m + 3.80

27


(b) Solve the system of equations using algebra. Verify your solution by graphing the system or by substituting your solution back into the original equations. (If an answer does not exist, enter DNE.)

28


Solution:

2.0m + 4.5 = 2.2m +3.8

-2.0m 2.0m

4.5 = .2m + 3.8

-3.8 - 3.8

.7 = .2m

.2 .2

m = 3.5

29


Solution: N(m) = 2.00m + 4.50 B(m) = 2.20m + 3.80

Plug (3.5) in for m into either equation to solve for your output.

N(3.5) = 2.00(3.5) + 4.50

N(3.5) = 11.50

(3.5, 11.50)

30


(c) Explain what your answer in part (b) means. (3.5, 11.50)

31


(c) Solution: The taxi fare in both New York and Baltimore is $11.50 for a trip 3.5 miles beyond the flag drop (assuming five minutes of waiting time).

32

Solving a System Using Graphs

33

Solving a System Using Graphs

One common approach to solving a system of equations is to graph all of the equations simultaneously and find the point of intersection. We do this in next Example.

34

Example 2 – Solutioncont’d

Figure 2.16

35

Example 2 – Solution

The graphs have been restricted to domains of x 0 because the number of miles traveled must be nonnegative to make sense.

The two lines appear to intersect at (2.5, 10).

We estimate this to be the solution of the system; however, we can use algebra to verify if this is an exact solution.

cont’d

36

Breakeven Point

37

Substitution Method: Breakeven Point

Business Applications of Linear Systems

Business analysts are often interested in profit, revenue, and costs of a company.

In evaluating a business plan, it is important to know at what sales level revenue and costs are expected to be equal. This sales level is referred to as the break-even point and is the point at which the company begins to turn a profit.

Any business has two types of costs: fixed costs and variable costs.

38


Fixed costs are those that remain constant regardless of production levels. For example, building rent, product research and development, and advertising are usually fixed costs.

Variable costs are those that vary with the level of production. For example, raw materials and production-line worker wages are variable costs.

39

Example 5 – Determining a Break-Even Point

An artisan wants to sell her handmade craft angels online. She estimates her material cost for each angel to be $3.50. As of January 2007, the online merchant craft mall. com charged a $14.95 per month fee for a Premier account featuring up to 25 products.

Comparing her craft to similar crafts on the market, the artisan estimates she can sell the craft angel for $9.95. How many angels will she have to sell each month to break even? At that production level, what will be her production cost, revenue, and profit?

40


The cost equation for the craft angels is the sum of the variable cost, $3.50 per angel, and the fixed cost, $14.95. Let a be the number of angels sold in a month.

The cost equation is

The revenue equation is

We want to determine when her revenue will equal her cost.

41


In other words, we want to find the value of a such that R(a) = C(a). Graphing the two functions simultaneously results in the graphs shown in Figure 2.18.

Figure 2.18

cont’d

42


It appears the graphs intersect near (2.3, 23) at the intersection point, R(a) = C(a). We can find the exact point of intersection by using the substitution method.

cont’d

43


We evaluate R(a) at a = 2.318 and determine that

Thus the break-even point is roughly (2.318, 23.06).

In the context of the problem, though, it does not make sense to talk about 2.318 angels. So we conclude she must sell 3 angels per month to cover her costs.

cont’d

44


The cost to produce and advertise 3 angels is

The revenue from the sale of 3 angels is

She will profit $4.40 if she sells 3 angels

cont’d

45

Solving a System of Three or More Equations

46

Solving a System of Three or More Equations

For a system of three or more linear equations to have a solution, all of the lines must intersect at the same point.

The fact that two lines intersect at a point (a, b) does not ensure that the third line will intersect the first two lines at the same point.

47

Example 11 – Solving a System of Three Equations

Solve the following system of equations.

Solution:

We will find the point of intersection of the first two equations and then check to see if the point is a solution to the third equation.

48


The first equation may be written as y = 2x – 5. Substituting this value in for y in the second equation and solving for x yields

cont’d

49


We substitute this value of x into y = 2x – 5 and solve.

The point of intersection of the first two lines is (3, 1). We will check to see if this point satisfies the third equation.

cont’d

50


Since the resultant statement is true, the solution to the system of equations is x = 3 and y = 1. (A false statement would have shown that the system was inconsistent.) We confirm the result in Figure 2.21.

cont’d

Figure 2.21

Copyright © Cengage Learning. All rights reserved. Systems of Linear Equations SECTION 2.4.

Documents

Transcript of Copyright © Cengage Learning. All rights reserved. Systems of Linear Equations SECTION 2.4.