Slide 1- 1 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley.

Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Systems of Linear Equations and Problem Solving

8.1 Systems of Equations in Two Variables8.2 Solving by Substitution or Elimination8.3 Solving Applications: Systems of Two

Equations8.4 Systems of Equations in Three Variables8.5 Solving Applications: Systems of Three

Equations8.6 Elimination Using Matrices8.7 Determinants and Cramer’s Rule 8.8 Business and Economic Applications

8


Systems of Equations in Two Variables

Translating

Identifying Solutions

Solving Systems Graphically

8.1


System of EquationsA system of equations is a set of two or more equations, in two or more variables, for which a common solution is sought.


Solution

T-shirt Villa sold 52 shirts, one kind at $8.25 and another at $11.50 each. In all, $464.75 was taken in for the shirts. How many of each kind were sold? Set up the equations but do not solve.

1. Familiarize. To familiarize ourselves with this problem, guess that 26 of each kind of shirt was sold. The total money taken in would be

26 $8.25 26 $11.50 $513.50

The guess is incorrect, now turn to algebra.

Example


2. Translate. Let x = the number of $8.25 shirts and y = the number of $11.50 shirts.

We have the following system of equations: 52,

8.25 11.50 464.75.

x y

x y

x + y = 52

8.25x + 11.50y = 464.75

Kind of Shirt

$8.25 shirt

$11.50 shirt

Total

Number sold

x y 52

Price $8.25 $11.50

Amount $8.25x $11.50y $464.75



A solution of a system of two equations in two variables is an ordered pair of numbers that makes both equations true.


Solution

Determine whether (1, 5) is a solution of the system

x – y = – 4

1 – 5 – 4– 4 = – 4 TRUE

4,

2 7.

x y

x y

2x + y = 7

2(1) + 5 77 = 7 TRUE

The pair (1, 5) makes both equations true, so it is a solution of the system.

Example


Solving Systems Graphically

One way to solve a system of two equations is to graph both equations and identify any points of intersection. The coordinates of each point of intersection represent a solution of that system.


,1

5x y

x y

Solve the system graphically.

Solution

It appears that (3, 2) is the solution. A check by substituting into both equations shows that (3, 2) is indeed the solution.

Example

Graph both equations.

x – y = 1

x + y = 5

(3, 2)



Solution

Example


,2 3

2 1y

y x

x

The lines have the same slope and different y-intercepts, so they are parallel. The system has no solution.



Solution

Example


3 6,

2 6 12

x y

y x

The same line is drawn twice. Any solution of one equation is a solution of the other. There is an infinite number of solutions. The solution set is ( , ) | 3 6 .x y x y


When we graph a system of two linear equations in two variables, one of the following three outcomes will occur.

1. The lines have one point in common, and that point is the only solution of the system. Any system that has at least one solution is said to be consistent.

2. The lines are parallel, with no point in common, and the system has no solution. This type of system is called inconsistent.

3. The lines coincide, sharing the same graph. This type of system has an infinite number of solutions and is also said to be consistent.


When one equation in a system can be obtained by multiplying both sides of another equation by a constant, the two equations are said to be dependent. If two equations are not dependent, they are said to be independent.


Solving by Substitution or Elimination

The Substitution Method

The Elimination Method

8.2


The Substitution MethodAlgebraic (nongraphical) methods for solving systems are often superior to graphing, especially when fractions are involved. One algebraic method, the substitution method, relies on having a variable isolated.


Solution

2,

1.

x y

y x

Solve the system (1)

(2)

The equations are numbered for reference.

Equation (2) says that y and x – 1 name the same number. Thus we can substitute x – 1 for y in equation (1):

x + y = 2

x + (x – 1) = 2

Equation (1)

Substituting x – 1 for y

We solve the last equation:

Example


Solution (continued)

x + (x – 1) = 2

Now return to the original pair of equations and substitute 3/2 for x in either equation so that we can solve for y

2x = 3

x = . 3

2

Equation (2)

Substituting 3/2 for x

1y x y = 3/2 – 1

y = . 1

2

2x – 1 = 2



3 1, .

2 2

We obtain the ordered pair

We can plug the ordered pair into the original pair of equations to check that it is the solution.


Solution

3 5,

2 3 7.

x y

x y


(2)

First, select an equation to solve for one variable. To isolate y, subtract 3x from both sides of equation (1):

3 5

5 3 .

x y

y x

(1)

(3)

Next, proceed as in the last example, by substituting:

Example



5 3y x

Equation (2)

Substituting 5 – 3x for y

We can substitute 2 for x in either equation (1), (2), or (3). It is easiest to use (3) because it has already been solved for y:

2x – 3y = 7

y = –1.

2x – 3(5 – 3x) = 7

2x – 15 + 9x = 7

11x = 22x = 2.

y = 5 – 3(2)



We obtain the ordered pair (2, –1).



Solution


(2)

We can substitute 7y + 2 for x in equation (1) and solve:

7y + 2 = 7y + 5

When the y terms drop out, the result is a contradiction. We state that the system has no solution.

7 5,

7 2.

x y

x y

2 = 5.

Substituting 7y + 2 for x

Subtracting 7y from both sides

Example


The Elimination Method

The elimination method for solving systems of equations makes use of the addition principle: If a = b, then a + c = b + c.


0x + 2y = 14.

Solution

5,

9.

x y

x y


(2)

Note that according to equation (2), –x + y and 9 are the same number. Thus we can work vertically and add –x + y to the left side of equation (1) and 9 to the right side:

x + y = 5 (1)

–x + y = 9 (2)

Adding

Example



This eliminates the variable x, and leaves an equation with just one variable, y for which we solve:

2y = 14

y = 7

x + 7 = 5Substituting. We also could have used equation (2).

Next, we substitute 7 for y in equation (1) and solve for x:

x = –2



We obtain the ordered pair (–2, 7).



Solution

2 6,

1 11.

4 3

x y

x y


(2)

To clear the fractions in equation (2), we multiply both sides of equation (2) by 12 to get equation (3):

1 112 12(1)

4 3

3 4 12.

x y

x y

(3)

Example



2 6,

3 4 12.

x y

x y

Now we solve the system (1)

(3)

Notice that if we add equations (1) and (3), we will not eliminate any variables. If the –2y in equation (1) were changed to –4y, we would. To accomplish this change, we multiply both sides of equation (1) by 2:

2x – 4y = –12 3x + 4y = 12

5x + 0y = 0x = 0.

Adding

Solving for x

(3)

Multiply eqn. (1) by 2



Then 3(0) + 4y = 12 Substituting 0 for x in equation (3)

4y = 12

y = 3

We obtain the ordered pair (0, 3).



Solution

2 3 2,

4 6 4.

x y

x y


(2)

Multiply equation (1) by 24x – 6y = 4– 4x + 6y = –4

Add 0 = 0

Note that what remains is an identity. Any pair that is a solution of equation (1) is also a solution of equation (2). The equations are dependent and the solution set is infinite:

( , ) | 2 3 2 , or equivalently ( , ) | 4 6 4 .x y x y x y x y

Example


Rules for Special CasesWhen solving a system of two linear equations in two variables:

1. If we obtain an identity such as 0 = 0, then the system has an infinite number of solutions. The equations are dependent and, since a solution exists, the system is consistent.

2. If we obtain a contradiction such as 0 = 7, then the system has no solution. The system is inconsistent.


Solving Applications: Systems of Two Equations

Total-Value and Mixture Problems

Motion Problems

8.3


Total-Value Problems

The next example involves two types of items, the quantity of each type bought, and the total value of the items. We refer to this type of problem as a total-value problem.


Solution

T-shirt Villa sold 52 shirts, one kind at $8.25 and another at $11.50 each. In all, $464.75 was taken in for the shirts. How many of each kind were sold?

1. Familiarize. To familiarize ourselves with this problem, guess that 26 of each kind of shirt was sold. The total money taken in would be

26 $8.25 26 $11.50 $513.50


Example


2. Translate. Let x = the number of $8.25 shirts and y = the number of $11.50 shirts.

We have the following system of equations: 52,

8.25 11.50 464.75.

x y

x y

x + y = 52

8.25x + 11.50y = 464.75

Kind of Shirt

$8.25 shirt

$11.50 shirt

Total

Number sold

x y 52

Price $8.25 $11.50

Amount $8.25x $11.50y $464.75


3. Carry out. We are to solve the system of equations 52,

8.25 11.50 464.75.

x y

x y

Where x is the number of $8.25 shirts and y is the number of $11.50 shirts. We can eliminate x by multiplying both sides of equation (1) by –8.25 and adding them to the corresponding sides of equation (2):

(1)

(2)

3.25y = 35.75

8.25x + 11.50y = 464.75

–8.25x – 8.25y = – 429

y = 11.


To find x, we substitute 11 for y in equation (1) and then solve for x:

x + 11 = 52x = 41

We obtain (41, 11), or x = 41 and y = 11.

4. Check. Recall x is the number of $8.25 shirts and y is the number of $11.50 shirts.

Number of shirts: x + y = 41 + 11 = 52

Money taken in: $8.25x + $11.50y

= $8.25(41) + $11.50(11) = 464.75

The numbers check.


5. State. T-shirt Villa sold 41 $8.25 shirts and 11 $11.50 shirts.


Mixture Problems

The next example is similar to the last example. Note that in each case, one of the equations in the system is a simple sum while the other equation represents a sum of products. We refer to this type of problem as a mixture problem.


Problem-Solving Tip

When solving a problem, see if it is patterned or modeled after a problem that you have already solved.


Solution

An employee at a small cleaning company wishes to mix a cleaner that is 30% acid and another cleaner that is 50% acid. How many liters of each should be mixed to get 20 L of a solution that is 35% acid?

1. Familiarize. We make a drawing and then make a guess to gain familiarity with the problem.

+

30% acid 50% acid 35% acid

=t liters

f liters

20 liters

Example


To familiarize ourselves with this problem, guess that 10 liters of each are mixed. The resulting mixture will be the right size but we need to check the strength:

0.30(10) 0.50(10) 8 L.


That is 8 L of acid for our guess but we want 0.35(20) = 7 L.


2. Translate. Let t = the number of liters of the 30% solution and f = the number of liters of the 50% solution.

We have the following system of equations:20,

0.30 0.50 7.

t f

t f

t + f = 20

0.30t + 0.50f = 7

First Solution

Second Solution

Mixture

Number of Liters

t f 20

Percentage of Acid

30% 50% 35%

Amount of Acid

0.30t 0.50f 7


3. Carry out. We are to solve the system of equations: 20,

0.30 0.50 7.

t f

x y

We can eliminate f by multiplying both sides of equation (1) by –0.5 and adding them to the corresponding sides of equation (2):

(1)

(2)

–0.20t = –3

0.30t + 0.50f = 7

–0.50t – 0.50f = –10

t = 15.


To find f, we substitute 15 for t in equation (1) and then solve for f:

15 + f = 20f = 5

We obtain (15, 5), or t = 15 and f = 5.

4. Check. Recall t is the number of liters of 30% solution and f is the number of liters of 50% solution.

Number of liters: t + f = 15 + 5 = 20

Amount of Acid: 0.30t + 0.50f

= 0.30(15) + 0.50(5) = 7

The numbers check.


5. State. The employee should mix 15 L of the 30% solution with 5 L of the 50% solution to get 20 L of a 35% solution.


Motion Problems

The next example deals with distance, speed (rate), and time. We refer to this type of problem as a motion problem.


Distance, Rate, and Time EquationsIf r represents rate, t represents time, and d represents distance, then

, , and .d d

d rt r tt r


Solution

Alex’s motorboat took 4 hr to make a trip downstream with a 5-mph current. The return trip against the same current took 6 hr. Find the speed of the boat in still water.

1. Familiarize. Note that the current speeds up the boat when going downstream, but slows down the boat when going upstream. For our guess, suppose that the speed of the boat with no current is 20 mph. The boat would then travel 20 + 5 = 25 mph downstream and 20 – 5 = 15 mph upstream.

Example


In 4 hr downstream the boat would travel 4(25) = 100 mi. In 6 hr upstream the boat would travel 6(15) = 90 mi. So our guess of 20 mph is incorrect.

Let r = the rate of the boat in still water. Then r + 5 = the boat’s speed downstream, and r – 5 = the boat’s speed upstream.

2. Translate.


Use Distance = (Rate)(Time).

We have the following system of equations:

( 5)4,

( 5)6.

d r

d r

d = (r + 5)4

d = (r – 5)6

Distance Rate Time

Downstream d r + 5 4

Upstream d r – 5 6


3. Carry out. We solve the system by substitution:(r + 5)4 = (r – 5)6

4r + 20 = 6r – 30

50 = 2r

25 = r

4. Check. When r = 25 mph, the speed downstream is 30 mph and the speed upstream is 20 mph. The distance downstream is 30(4) = 120 mi and the distance upstream is 20(6) = 120 mi, so we have a check.

5. State. The speed of the boat in still water is 25 mph.


Tips for Solving Motion Problems1. Draw a diagram using an arrow or arrows to

represent distance and the direction of each object in motion.

2. Organize the information in a chart.

3. Look for times, distances, or rates that are the same. These often can lead to an equation.

4. Translating to a system of equations allows for the use of two variables.

5. Always make sure that you have answered the question asked.


Systems of Equations in Three Variables


Solving Systems in Three Variables

Dependency, Inconsistency, and Geometric Considerations

8.4



A solution of a system of three equations in three variables is an ordered triple (x, y, z) that makes all three equations true.

A linear equation in three variables is an equation equivalent to one in the form Ax + By + Cz = D, where A, B, C, and D are real numbers. We refer to the form Ax + By + Cz = D as standard form for a linear equation in three variables.


Determine whether (2, –1, 3) is a solution of the system 4,

2 2 3,

4 2 3.

x y z

x y z

x y z

Example


Solution We substitute (2, –1, 3) into the three equations, using alphabetical order:

x + y + z = 4

2 + (–1) + 3 44 = 4 3 = 3TRUE TRUE

The triple makes all three equations true, so it is a solution of the system.

2x – 2y – z = 3

2(2) – 2(–1) – 3 3

– 4x + y + 2z = –3

– 4(2) + (–1) + 2(3) –3 –3 = –3 TRUE


Solving Systems in Three Variables

The elimination method allows us to manipulate a system of three equations in three variables so that a simpler system of two equations in two variables is formed. Once that simpler system has been solved, we can substitute into one of the three original equations and solve for the third variable.


Solution

Solve the following system of equations:

6,

2 2,

3 8.

x y z

x y z

x y z

We select any two of the three equations and work to get one equation in two variables. Let’s add equations (1) and (2):

(1)

(2)

(3)

6

2 2

x y z

x y z

(1)

(2)

(4)2x + 3y = 8Adding to eliminate z

Example


Next, we select a different pair of equations and eliminate the same variable. Let’s use (2) and (3) to again eliminate z.

2 2

3 8

x y z

x y z

(5) x – y + 3z = 8

Multiplying equation (2) by 3

4x + 5y = 14.

3x + 6y – 3z = 6

Now we solve the resulting system of equations (4) and (5). That will give us two of the numbers in the solution of the original system,

4x + 5y = 14

2x + 3y = 8(5)

(4)


We multiply both sides of equation (4) by –2 and then add to equation (5):

Substituting into either equation (4) or (5) we find that x = 1.

4x + 5y = 14–4x – 6y = –16,

–y = –2 y = 2

Now we have x = 1 and y = 2. To find the value for z, we use any of the three original equations and substitute to find the third number z.


Let’s use equation (1) and substitute our two numbers in it:

We have obtained the ordered triple (1, 2, 3). It should check in all three equations.

x + y + z = 6

1 + 2 + z = 6z = 3.

The solution is (1, 2, 3).


Solving Systems of Three Linear Equations To use the elimination method to solve systems of three linear equations:

1. Write all the equations in standard form Ax + By+ Cz = D.

2. Clear any decimals or fractions.

3. Choose a variable to eliminate. Then select two of the three equations and work

to get one equation in which the selected variable is eliminated.


Solving Systems of Three Linear Equations (continued)

4. Next, use a different pair of equations and eliminate the same variable that you

did in step (3).

5. Solve the system of equations that resulted from steps (3) and (4).

6. Substitute the solution from step (5) into one of the original three equations and solve for

the third variable. Then check.


Dependency, Inconsistency, and Geometric Considerations

The graph of a linear equation in three variables is a plane. Solutions are points common to the planes of each system. Since three planes can have an infinite number of points in common or no points at all in common, we need to generalize the concept of consistency in three dimensions.


Consistency

A system of equations that has at least one solution is said to be consistent.

A system of equations that has no solution is said to be inconsistent.


Solution

Solve the following system of equations: 2 2,

2 5,

1.

y z

x y z

x y z

The variable x is missing in equation (1). Let’s add equations (2) and (3) to get another equation with x missing:

(1)

(2)

(3)

2 5

1

x y z

x y z

(2)

(3)

(4)y + 2z = 6 Adding

Example


Equations (1) and (4) form a system in y and z. Solve as before:

Multiplying equation (1) by –1

0 = 4.

Since we ended up with a false equation, or contradiction, we know that the system has no solution. It is inconsistent.

y + 2z = 6

y + 2z = 2 –y – 2z = –2

y + 2z = 6


Recall that when dependent equations appeared in Section 8.1, the solution sets were always infinite in size and were written in set-builder notation. There, all systems of dependent equations were consistent. This is not always the case for systems of three or more equations. The following figures illustrate some possibilities geometrically.


Solving Applications: Systems of Three Equations

Applications of Three Equations in Three Unknowns

8.5


Solution

The sum of three numbers is 6. The first number plus twice the second, minus the third is 2. The first minus the second, plus three times the third is 8.

1. Familiarize. There are three statements involving the same three numbers. Let’s label these numbers x, y, and z.

Example


2. Translate. We can translate directly as follows.

The sum of three numbers is 6.

x + y + z = 6

The first number plus twice the second, minus the third is 2.

x + 2y – z = 2

The first minus the second, plus three times the third is 8.

x – y + 3z = 8


We now have the system of three equations: 6,

2 2,

3 8.

x y z

x y z

x y z

3. Carry out. We need to solve the system of equations. Note that we found the solution, (1, 2, 3) in an example in Section 3.4.

5. State. The three numbers are 1, 2, and 3.

4. Check. The check is left to the student.


Solution

In triangle ABC, the measure of angle B is three times the measure of angle A. The measure of angle C is 60o greater than twice the measure of angle A. Find the measure of each angle.

1. Familiarize. We can make a sketch and label the angles A, B, and C. Recall that the measures of the angles in any triangle add to 180o.

AB

C

Example


2. Translate. This geometric fact about triangles gives us one equation:

A + B + C = 180.

B = 3A

C = 60 + 2A

Angle C is 60o greater than twice the measure of A.

Angle B is three times the measure of angle A.


We now have the system of three equations:

180,

3 ,

60 2

A B C

B A

C A

3. Carry out. We need to solve the system of equations. Through the techniques of Section 3.4 we find the solution to be (20, 60, 100).

5. State. The angles in the triangle measure 20o, 60o, and 100o.

4. Check. The check is left to the student.


Elimination Using Matrices

Matrices and Systems

Row-Equivalent Operations

8.6


In solving systems of equations, we perform computations with the constants. The variables play no important role until the end.

3 5,

2 3 7.

x y

x y

simplifies to 3 1 5

2 –3 7

if we do not write the variables, the operation of addition, and the equals signs.

For example, the system


Matrices and SystemsIn the previous slide, we have written a rectangular array of numbers. Such an array is called a matrix (plural, matrices). We ordinarily write brackets around matrices. The following are examples of matrices:

1 1 6 9 1/ 2 01 2

, 4 , 4 3 1 7 67 5

0 6 5 9 1 2

The individual numbers are called elements or entries.


The rows of a matrix are horizontal, and the columns are vertical.

1 9 0

4 1 6

6 9 2

column 1 column 2 column 3

row 1

row 2

row 3


Solution

3 7,

3 1.

x y

x y

Use matrices to solve the system.

We write a matrix using only coefficients and constants, listing x-coefficients in the first column and y-coefficients in the second. Note that in each matrix a dashed line separates the coefficients from the constants:

3 1 7

1 3 1

Example


Our goal is to transform

The variables x and y can then be reinserted to form equations from which we can complete the solution.

We do calculations that are similar to those that we would do if we wrote the entire equations.

3 1 7

3 9 3

New Row 2 = 3(Row 2 from above)

3 1 7

1 3 1

into the form .0

a b c

d e


If we now reinsert the variables, we have

3 7,

10 10.

x y

y

Now proceed as before, solving equation (2) for y:

y = 1.

(1)

(2)

10y = 10,

3 1 7

0 10 10

New Row 2 = Row 1 + 3(Row 2)


Next, we substitute 1 in for y in equation (1):

The solution is (2, 1). The check is left to the student.

3x + 1 = 7

3x + y = 7

3x = 6

x = 2.


2 8,

1,

2 2.

x y z

x y z

x y z

Use matrices to solve the system.

SolutionWe write a matrix using only coefficients and constants.

2 1 1 8

1 1 1 1

1 2 1 2

Example


Our goal is to transform

into the form

2 1 1 8

1 1 1 1

1 2 1 2

0 .

0 0

a b c d

e f g

h i

Interchange Row 1 and Row 21 1 1 1

2 1 1 8

1 2 1 2

1 1 1 1

0 3 3 6

1 2 1 2

New Row 2 = –2(Row 1) + Row 2


1 1 1 1

0 1 1 2

0 1 2 1

1 1 1 1

0 1 1 2

0 0 3 3

1,

2,

3 3.

x y z

y z

z

If we now reinsert the variables, we have(1)

(2)

(3)

New Row 3 = –(Row 2) + Row 3

New Row 2 = (1/3)(Row 2)

New Row 3 = Row 1 + Row 3


Now proceed as before solving equation (3) for z:

z = –1.

Substitute the z value into equation (2) to find that y = 1. Then substitute the z and y values into equation (1) to find x = 3. The solution is (3, 1, –1).

The check is left to the student.


Row-Equivalent Operations

Each of the following row-equivalent operations produces a row-equivalent matrix:

a) Interchanging any two rows.

b) Multiplying all elements of a row by a nonzero constant.

c) Replacing a row with the sum of that row and a multiple of another row.


Determinants and Cramer’s Rule

Determinants of 2 x 2 Matrices

Cramer’s Rule: 2 x 2 Systems

Cramer’s Rule: 3 x 3 Systems

8.7


Determinants of 2 x 2 Matrices

When a matrix has m rows and n columns, it is called an “m by n” matrix. Thus its dimensions are denoted by m x n. If a matrix has the same number of rows and columns, it is called a square matrix. Associated with every square matrix is a number called its determinant, defined as follows for 2 x 2 matrices.


2 x 2 DeterminantsThe determinant of a two-by-two matrix

and is defined as follows:

is denoted a c a c

b d b d

.a c

ad bcb d


Evaluate: 3 2.

7 1

Solution

3 2

7 1

= 3(1) – 7(–2) = 17.

Multiply and subtract as follows:

Example


Cramer’s Rule: 2 x 2 Matrices

1 1 1

2 2 2

,

,

a x b y c

a x b y c

Using the elimination method, we can show that the solution to the system

is1 2 2 1 1 2 2 1

1 2 2 1 1 2 2 1

and .c b c b a c a c

x ya b a b a b a b

These fractions can be rewritten using determinants.


Cramer’s Rule: 2 x 2 SystemsThe solution of the system

if it is unique, is given by

1 1 1

2 2 2

,

,

a x b y c

a x b y c

1 1 1 1

2 2 2 2

1 1 1 1

2 2 2 2

, .

c b a c

c b a cx y

a b a b

a b a b


Cramer’s Rule: 2 x 2 Systems (continued)

These formulas apply only if the denominator is not 0. If the denominator is 0, then one of two things happens:

1.If the denominator is 0 and the numerators are also 0, then the equations in the system are dependent.

2.If the denominator is 0 and at least one numerator is not 0, then the system is inconsistent.


Solution

6 2,

2 3 2.

x y

x y

Solve using Cramer’s rule:

We have

2 1

2 36 1

2 3

x

2( 3) ( 2)(1)

6( 3) 2(1)

4 1

20 5

Example



and

6 2

2 26 1

2 3

y

6( 2) 2(2)

6( 3) 2(1)

16 4

.20 5

The solution is (1/5, 4/5).


3 x 3 DeterminantsThe determinant of a three-by-three matrix is defined as follows:

1 1 1

2 2 1 1 1 1

2 2 2 1 2 3

3 3 3 3 2 2

3 3 3

a b cb c b c b c

a b c a a ab c b c b c

a b c

Subtract. Add.


Solution

Evaluate: 1 0 3

2 7 2 .

1 3 4

= 1(34) – 2(–9) + (–1)(–21)

= 73.

1 0 37 2 0 3 0 3

2 7 2 (1) (2) ( 1)3 4 3 4 7 2

1 3 4

Example


Cramer’s Rule: 3 x 3 SystemsThe solution of the system

can be found by using the following determinants:

1 1 1 1

2 2 2 2

3 3 3 3

,

,

a x b y c z d

a x b y c z d

a x b y c z d

1 1 1 1 1 1

2 2 2 2 2 2

3 3 3 3 3 3

, x

a b c d b c

D a b c D d b c

a b c d b c


Cramer’s Rule: 3 x 3 Systems(continued)

If a unique solution exists, it is given by

1 1 1 1 1 1

2 2 2 2 2 2

3 3 3 3 3 3

, .y z

a d c a b d

D a d c D a b d

a d c a b d

, , .yx zDD D

x y zD D D


Solution

2 8,

1,

2 2.

x y z

x y z

x y z

Solve using Cramer’s rule:

Compute the determinants:

2 1 1 8 1 1

1 1 1 9, 1 1 1 27

1 2 1 2 2 1xD D

Example



2 8 1 2 1 8

1 1 1 9, 1 1 1 9.

1 2 1 1 2 2y zD D

27 93, 1,

9 99

1.9

yx

z

DDx y

D DD

zD

Then

The solution is (3, 1, –1). The check is left to the student.


Business and Economic Applications

Break-Even Analysis

Supply and Demand

8.8


Break-Even AnalysisWhen a company manufactures x units of a product, it spends money. This is total cost and can be thought of as a function C, where C(x) is the total cost of producing x units. When a company sells x units of the product, it takes in money. This is total revenue and can be thought of as a function R, where R(x) is the total revenue from the sale of x units.


Break-Even Analysis

Total profit is the money taken in less the money spent, or total revenue minus total cost. Total profit from the production and sale of x units is a function P given by

Profit = Revenue – Cost, or

P(x) = R(x) – C(x).

(continued)


Cost

There are two types of costs. Costs which must be paid whether a product is produced or not, are called fixed costs. Costs that vary according to the amount being produced are called variable costs. The sum of the fixed cost and variable cost gives the total cost.


A specialty wallet company has fixed costs that are $2,400. Each wallet will cost $2 to produce (variable costs) and will sell for $10. a) Find the total cost C(x) of producing x wallets.

b) Find the total revenue R(x) from the sale of x wallets.

c) Find the total profit P(x) from the production and sale of x wallets.

d) What profit will the company realize from the production and sale of 500 wallets?

e) Graph the total-cost, total-revenue, and total-profit functions. Determine the break-even point.

Example


Solution

a) Total cost is given by

C(x) = (Fixed costs) plus (Variable costs)

C(x) = 2,400 + 2x.

where x is the number of wallets produced.

b) Total revenue is given by

R(x) = 10x $10 times the number of wallets sold.

c) Total profit is given byP(x) = R(x) – C(x) = 10x – (2,400 + 2x)

= 8x – 2,400.

A specialty wallet company has fixed costs that are $2,400. Each wallet will cost $2 to produce (variable costs) and will sell for $10.


Solution

d) Total profit will be

P(500) = 8(500) – 2,400

= 4,000 – 2,400

= $1,600.

e) The graphs of each of the three functions are shown on the next slide. R(x), C(x), and P(x) are all in dollars.

A specialty wallet company has fixed costs that are $2,400. Each wallet will cost $2 to produce (variable costs) and will sell for $10.


e)

0 50 100 150 200 250 300 350 400 450 500 550

500

1,5001,000

2,5002,000

3,000

4,0003,500

R(x) = 10x

C(x) = 2400 + 2x

P(x) = 8x – 2400

-2500

Break-even point

Loss

Gain

Wallets sold


Gains occur where the revenue is greater than the cost. Losses occur where the revenue is less than the cost. The break-even point occurs where the graphs of R and C cross. Thus to find the break-even point , we solve the system:

( ) 10 ,

( ) 2,400 2 .

R x x

C x x

Using substitution we find that x = 300. The company will break even if it produces and sells 300 wallets and takes in a total of R(300) = $3,000 in revenue. Note that the break-even point can also be found by solving P(x) = 0.


Supply and Demand

As the price of a product varies, the amount sold varies. Consumers will demand less as price goes up. Sellers will supply more as the price goes up.


Supply and Demand

Supply

Demand

Equilibrium point

Price

Qua

ntit

y

The point of intersection is called the equilibrium point. At that price, the amount that the seller will supply is the same amount that the consumer will buy.


Solution

( ) 3000 80 ,

( ) 120 10 .

D p p

S p p

Find the equilibrium point for the demand and supply functions given.

Since both demand and supply are quantities and they are equal at the equilibrium point, we rewrite the system as

(1)

(2)

(1)

(2)

3000 80 ,

120 10 .

q p

q p

Example


Solving using substitution we find the equilibrium price is $32. To find the quantity, we substitute $32 into either equation D(p) or S(p). We use S(p):

(32) 120 10(32) 440.S

Thus, the equilibrium quantity is 440 units, and the equilibrium price is $32.


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