5.1 - 1 Copyright © 2013, 2009, 2005 Pearson Education, Inc. 1 5 Systems and Matrices Copyright ©...

5.1 - 1Copyright © 2013, 2009, 2005 Pearson Education, Inc. 1

5Systems and Matrices

Copyright © 2013, 2009, 2005 Pearson Education, Inc.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 2

5.1 Systems of Linear Equations

• Linear Systems• Substitution Method• Elimination Method• Special Systems• Applying Systems of Equations• Solving Linear Systems with Three Unknowns (Variables)

• Using Systems of Equations to Model Data


Linear Systems

The definition of a linear equation given in Chapter 1 can be extended to more variables. Any equation of the form

1 1 2 2 ,n na x a x a x b

for real numbers a1, a2, …, an (all nonzero) and b, is a linear equation or a first-degree equation in n unknowns.


Linear Systems

A set of equations is called a system of equations. The solutions of a system of equations must satisfy every equation in the system. If all the equations in a system are linear, the system is a system of linear equations, or a linear system.


Linear Systems

The possible graphs of alinear system in two unknowns are as follows.1. The graphs intersect at exactly one point, which gives the (single) ordered-pair solution of the system. The system is consistent and the equations are independent.


Linear Systems

2. The graphs are parallel lines, so there is no solution and the solution set is Ø. The system is inconsistent and the equations are independent.


Linear Systems

3. The graphs are the same line, and there is an infinite number of solutions. The system is consistent and the equations are dependent.


Substitution Method

In a system of two equations with two variables, the substitution method involves using one equation to find an expression for one variable in terms of the other, and then substituting into the other equation of the system.


Example 1 SOLVING A SYSTEM BY SUBSTITUTION

Solve the system.

3 2 11x y 3x y

(1)

(2)Solution

Begin by solving one of the equations for one of the variables. We solve equation (2) for y.

3x y (2)

3y x Add x.



Now replace y with x + 3 in equation (1), and solve for x.

3 2 11x y (1)

3 2( 113)xx Let y = x + 3 in (1).Note the careful

use of parentheses.

3 2 6 11x x Distributive property

5 6 11x Combine like terms.

5 5x Subtract 6.

1x Divide by 5.


Replace x with 1 in equation (3) to obtain y = 1 + 3 = 4. The solution of the system is the ordered pair (1, 4). Check this solution in both equations (1) and (2).


3 2 11x y (1)

Check

3yx (2)?

3( ) 2( 14) 11

11 11 True

?

31 4 3 3 True

True statements result, confirming the solution set is {(1, 4)}.


Elimination Method

Another way to solve a system of two equations, called the elimination method, uses multiplication and addition to eliminate a variable from one equation. To eliminate a variable, the coefficients of that variable in the two equations must be additive inverses. To achieve this, we use properties of algebra to change the system to an equivalent system, one with the same solution set. The three transformations that produce an equivalent system are listed next.


Transformations of a Linear System1. Interchange any two equations of the

system.2. Multiply or divide any equation of the

system by a nonzero real number.3. Replace any equation of the system by

the sum of that equation and a multiple of another equation in the system.


Example 2 SOLVING A SYTEM BY ELIMINATION

Solve the system.

One way to eliminate a variable is to use the second transformation and multiply both sides of equation (2) by – 3, to get an equivalent system.

Solution

3 4 1x y (1)

(2)2 3 12x y

3 4 1x y (1)

6 9 36x y Multiply (2) by – 3 (3)



(3)6 9 36x y

Now multiply both sides of equation (1) by 2, and use the third transformation to add the result to equation (3), eliminating x. Solve the result for y.

6 8 2x y Multiply (1) by 2

17 34y Add.

2y Solve for y.



(1)3 4 1x y

Let y = 2 in (1).

Multiply.

Substitute 2 for y in either of the original equations and solve for x.

3 4( ) 12x

3 8 1x

3 9x Add 8.

3x Divide by 3.



A check shows that (3, 2) satisfies both equations (1) and (2). Therefore, the solution set is {(3, 2)}.


Example 3 SOLVING AN INCONSISTENT SYSTEM

Solve the system.3 2 4x y

6 4 7x y

(1)

(2)

To eliminate the variable x, multiply both sides of equation (1) by 2 and add the result to equation (2).

Solution

6 4 8x y Multiply (1) by 2

6 4 7x y (2)

0 15 False.


Example 3 SOLVING AN INCONSISTENT SYSTEM

Since 0 = 15 is false, the system is inconsistent and has no solution. As suggested by the graph, this means that the graphs of the equations of the systemnever intersect. (The lines are parallel.) The solution set is Ø.


Example 4SOLVING A SYSTEM WITH INFINITELY MANY SOLUTIONS

Solve the system.

Solution

Divide both sides of equation (1) by 2, and add the result to equation (2).

4 2x y Divide (1) by 2.

8 2 4x y

4 2x y

(1)

(2)

4 2x y (2)

0 0 True.



The result, 0 = 0, is a true statement, which indicates that the equations of the original system are equivalent. Any ordered pair (x, y) that satisfies either equation will satisfy the system. Solve for y in equation (2).

4 2x y (2)

4 2y x



The solutions of the system can be written in the form of a set of ordered pairs (x, 4x + 2), for any real number x. Some ordered pairs in the solution set are (0, 4 • 0 + 2), or (0, 2), and (1, 4 • 1 + 2), or (1, 6), as well as (3, 14), and (– 2, – 6). As shown here, the equations of the original system are dependent and lead to the same straight-line graph. The solution set can be written {(x, 4x + 2)}.


Note In the algebraic solution for Example 4, we wrote the solution set with the variable x arbitrary. We could write the solution set with y arbitrary. 2

,4

yy

By selecting values for y and solving for x in this ordered pair, we can find individual solutions. Verify again that (0, 2) is a solution by letting y = 2 and solving for x to obtain 2 2

0.4


Applying Systems of Equations

Many applied problems involve more than one unknown quantity. Although some problems with two unknowns can be solved using just one variable, it is often easier to use two variables.

To solve a problem with two unknowns, we must write two equations that relate the unknown quantities. The system formed by the pair of equations can then be solved using the methods of this chapter.


Solving An Applied Problem By Writing A System of EquationsStep 1 Read the problem carefully until you understand

what is given and what is to be found.Step 2 Assign variables to represent the unknown values,

using diagrams or tables as needed. Write down what each variable represents.

Step 3 Write a system of equations that relates the unknowns.

Step 4 Solve the system of equations.Step 5 State the answer to the problem. Does it seem

reasonable?Step 6 Check the answer in the words of the original

problem.


Example 5 USING A LINEAR SYSTEM TO SOLVE AN APPLICATION

Salaries for the same position can vary depending on the location. In 2010, the average of the salaries for the position of Accountant I in San Diego, California, and Salt Lake City, Utah, was $45,091.50. The salary in San Diego, however, exceeded the salary in Salt Lake City by $5231. Determine the salary for the Accountant I position in San Diego and in Salt Lake City.



Solution

Step 1 Read the problem. We must find the salary of the Accountant I position in San Diego and in Salt Lake City.

Step 2 Assign variables. Let x represent the salary of the Accountant I position in San Diego and y represent the salary for the same position in Salt Lake City.



Solution

Step 3 Write a system of equations. Since the average of the salaries for the Accountant I position in San Diego and Salt Lake City was $45,091.50, one equation is as follows.

45,091.502

x y

Multiply both sides of this equation by 2 to clear the fraction and get an equivalent equation. 90,183x y (1)



The salary in San Diego exceeded the salary in Salt Lake City by $5231. Thus, x – y = 5231, which gives the following system of equations.

90,183x y (1)

5231x y (2)



90,183x y (1)

5231x y (2)

Step 4 Solve the system. To eliminate y, add the two equations.

2 95,414x Add.

47,707x Solve for x.

To find y, substitute 47,707 for x in equation (2).47,707 5231y Let x = 47,707 in (2).

42,476y Subtract 47,707.

42,476y Multiply by – 1.



Step 5 State the answer. The salary for the position of Accountant I was $47,707 in San Diego and $42,476 in Salt Lake City.

Step 6 Check. The average of $47,707 and $42,476 is

$47,707 $42,476$45,091.50

2

Also, $47,707 – $42,476 = $5231, as required.


Solving Linear Systems with Three Unknowns (Variables)

Earlier, we saw that the graph of a linear equation in two unknowns is a straightline. The graph of a linear equation in three unknowns requires a three-dimensional coordinate system. The three number lines are placed at right angles. The graph of a linear equation in three unknowns is a plane.


Solving Linear Equations with three Unknowns (Variables)

Some possible intersections of planes representing three equations in three variables are shown here.








Solving a SystemSolve a linear system with three unknowns as follows.Step 1 Eliminate a variable from any two of the equations.Step 2 Eliminate the same variable from a different pair of equations. Step 3 Eliminate a second variable using the resulting two equations in two variables to get an equation with just one variable whose value we can now determine. Step 4 Find the values of the remaining variables by substitution. Write the solution of the system as an ordered triple.


Example 6 SOLVING A SYSTEM OF THREE EQUATIONS WITH THREE VARIABLES

Solve the system.3 9 6 3x y z

2 2x y z 2x y z

(1)

(2)

(3)Solution Eliminate z by adding equations (2) and (3). (Step 1)

3 2 4x y (4)


Example 6

To eliminate z from another pair of equations, multiply both sides of equation (2) by 6 and add the result to equation (1). (Step 2)

12 6 6 12x y z Multiply (2) by 6.

3 9 6 3x y z 15 15 15x y

(1)

Make sure equation (5) has

the same two variables as equation (4).

(5)

SOLVING A SYSTEM OF THREE EQUATIONS WITH THREE VARIABLES


Example 6

15 10 20x y Multiply (4) by – 5.

15 15 15x y 5 5y

(5)

To eliminate x from equations (4) and (5), multiply both sides of equation (4) by –5 and add the result to equation (5). Solve the resulting equation for y. (Step 3)

Add.

1y Divide by 5.



Example 6

(4) with y = – 1

Using y = – 1 , find x from equation (4) by substitution. (Step 4)

)13 2( 4x

2x


Solve for x.


Example 6

(3) with x = 2, y = – 1( ) 22 1 z

1z

Substitute 2 for x and – 1 for y in equation (3) to find z.

Verify that the ordered triple (2, –1, 1) satisfies all three equations in the original system. The solution set is {(2, –1,1)}.


Solve for z.


Caution When eliminating a variable from any two equations to create equations like (4) and (5) in Example 6, be sure to eliminate the same variable from both equations. Otherwise, the result will include two equations that still have three variables, and no progress will have been made.


Example 7 SOLVING A SYSTEM OF TWO EQUATIONS WITH THREE VARIABLES

Solve the system.2 4x y z

3 4 9x y z (1)

(2)

Geometrically, the solution is the intersection of the two planes given by equations (1) and (2). The intersection of two different nonparallel planes is a line. Thus there will be an infinite number of ordered triples in the solution set, representing the points on the line of intersection.

Solution

To eliminate x, multiply both sides of equation (1) by –3 and add the result to equation (2). (Either y or z could have been eliminated instead.)



3 6 3 12x y z Multiply (1) by – 3.

3 4 9x y z (2)

7 7 21y z (3)

7 7 21z y Add 7y.

3z y Divide each term by – 7.

Solve this equation for z.



2 4x y z 2 4x y z

32 ( ) 4yx y

1x y

Substitute (– y + 3) for z.

Simplify.

Use parentheses around – y + 3.

This gives z in terms of y. Express x also in terms of y by solving equation (1) for x and substituting –y + 3 for z in the result.

(1)Solve for x.

The system has an infinite number of solutions. With y arbitrary, the solution set is of the form {(–y + 1, y, – y + 3)}.


Note Had we solved equation (3) in Example 7 for y instead of z, the solution would have had a different form but would have led to the same set of solutions. In that case we would have z arbitrary, and the solution set would be of the form {(z – 2, –z + 3, z)}.By choosing z = 2, one solution would be (0, 1, 2).


Modeling Data

Applications with three unknowns usually require solving a system of three equations. If we know three points on the graph, we can find the equation of a parabola in the form

by solving a system of three equations with three variables.

2y ax bx c


Example 8 USING MODELING TO FIND AN EQUATION THROUGH THREE POINTS

Find the equation of the parabola y = ax2 + bx + c that passes through the points (2, 4), (– 1, 1), and (– 2, 5).

Solution

Since the three points lie on the graph of the given equation y = ax2 + bx + c, they must satisfy the equation. Substituting each ordered pair into the equation gives three equations with three unknowns.


Example 8

24 ( (2) ) ,2a b c or 4 4 2a b c (1)

2( )1 (1 )1 ,a b c or 1 a b c (2)

2( )5 (2 )2 ,a b c or 5 4 2a b c (3)

USING MODELING TO FIND AN EQUATION THROUGH THREE POINTS


Example 8

4 4 2a b c (1)

1 a b c

To solve this system, first eliminate c using equations (1) and (2).

Multiply (2) by – 1.

3 3 3a b (4)



Example 8

1 a b c (2)

5 4 2a b c Multiply (3) by – 1.

4 3a b (5)

Now, use equations (2) and (3) to eliminate the same unknown, c.

Equation (5) must have the same two unknowns

as equation (4).



Example 8

3 3 3a b (4)

4 3a b

Add.1 4b

(5)

Solve the system of equations (4) and (5) in two unknowns by eliminating a.

14

b Divide by 4.



Example 8

1 a b Equation (4) divided by 3

114

a Let b = – ¼.

54

a Add ¼ .

Find a by substituting – ¼ for b in equation (4).



Finally, find c by substituting and in equation (2).

Example 8

1 a b c (2)

14

54

1 c

6

14

c Add.

54

a 14

b

5Let ,

4a

1.

4b

12

c Subtract .

64

The required equation is 2 14

5 1,

24y x x or

21.25 0.25 0.5.y x x



Example 9 SOLVING AN APPLICATION USING A SYSTEM OF THREE EQUATIONS

An animal feed is made from three ingredients: corn, soybeans, and cottonseed. One unit of each ingredient provides units of protein, fat, and fiber as shown in the table. How many units of each ingredient should be used to make a feed that contains 22 units of protein, 28 units of fat, and 18 units of fiber?

Corn Soybeans Cottonseed Total

Protein 0.25 0.4 0.2 22Fat 0.4 0.2 0.3 28Fiber 0.3 0.2 0.1 18



Step 1 Read the problem. We must determine the number of units of corn, soybeans, and cottonseed.

Solution

Step 2 Assign variables. Let x represent the number of units of corn, y the number of units of soybeans, and z the number of units of cottonseed.



Step 3 Write a system of equations. The total amount of protein is to be 22 units, so we use the first row of the table to write equation (1).

0.25 0.4 0.2 22.x y z (1)

(2)

We use the second row of the table to obtain 28 units of fat.

0.4 0.2 0.3 28x y z Finally, we use the third row of the table to obtain 18 units of fiber,

0.3 0.2 0.1 18x y z (3)



25 40 20 2200x y z (4)

(5)4 2 3 280x y z

3 2 180x y z (6)

Multiply equation (1) on both sides by 100, and equations (2) and (3) by 10 to get an equivalent system.



Step 4 Solve the system. Using the methods described earlier in this section, we find that x = 40, y = 15, and z = 30.

Step 5 State the answer. The feed should contain 40 units of corn, 15 units of soybeans, and 30 units of cottonseed.

Step 6 Check. Show that the ordered triple (40, 15, 30) satisfies the system formed by equations (1), (2), and (3).


Note Notice how the table in Example 9 is used to set up the equations of the system. The coefficients in each equation are read from left to right. This idea is extended in the next section, where we introduce solution of systems by matrices.

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