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A system of equations is a set of equations that involve the same variables. A solution of a system is an assignment of values for the variables that make each equation in the system true.

1.
College Algebra

Sixth Edition

James Stewart Lothar Redlin Saleem Watson

Sixth Edition

James Stewart Lothar Redlin Saleem Watson

2.
Systems of

5 Equations and

Inequalities

5 Equations and

Inequalities

3.
Chapter Overview

We have already seen how a real-world

situation can be modeled by an equation.

However, Many real-world situations have

too many variables to be modeled by

a single equation.

• For example, weather depends on

many variables, including temperature, wind speed,

air pressure, and humidity.

We have already seen how a real-world

situation can be modeled by an equation.

However, Many real-world situations have

too many variables to be modeled by

a single equation.

• For example, weather depends on

many variables, including temperature, wind speed,

air pressure, and humidity.

4.
Chapter Overview

So, to model (and forecast) the weather,

scientists use many equations, each

having many variables.

• Such systems of equations work together

to describe the weather.

So, to model (and forecast) the weather,

scientists use many equations, each

having many variables.

• Such systems of equations work together

to describe the weather.

5.
Chapter Overview

Systems of equations with hundreds,

or even thousands, of variables are also

used extensively in the air travel and

telecommunications industries to:

• Establish consistent airline schedules.

• Find efficient routing for telephone calls.

Systems of equations with hundreds,

or even thousands, of variables are also

used extensively in the air travel and

telecommunications industries to:

• Establish consistent airline schedules.

• Find efficient routing for telephone calls.

6.
Systems of

Linear Equations

5.1 in Two Variables

Linear Equations

5.1 in Two Variables

7.
Systems of Linear Equations

and Their Solutions

and Their Solutions

8.
Systems of Linear Equations and Their Solutions

A system of equations is a set of equations

that involve the same variables.

A solution of a system is an assignment

of values for the variables that makes each

equation in the system true.

• To solve a system means to find

all solutions of the system.

A system of equations is a set of equations

that involve the same variables.

A solution of a system is an assignment

of values for the variables that makes each

equation in the system true.

• To solve a system means to find

all solutions of the system.

9.
Systems of Linear Equations and Their Solutions

Here is an example of a system of

linear equations in two variables:

2 x y 5 Equation 1

x 4 y 7 Equation 2

Here is an example of a system of

linear equations in two variables:

2 x y 5 Equation 1

x 4 y 7 Equation 2

10.
Systems of Linear Equations and Their Solutions

We can check that x = 3 and y = 1

is a solution of this system.

Equation 1 Equation 2

2x – y = 5 x + 4y = 7

2(3) – 1 = 5 3 + 4(1) = 7

• The solution can also be written as

the ordered pair (3, 1).

We can check that x = 3 and y = 1

is a solution of this system.

Equation 1 Equation 2

2x – y = 5 x + 4y = 7

2(3) – 1 = 5 3 + 4(1) = 7

• The solution can also be written as

the ordered pair (3, 1).

11.
Systems of Linear Equations and Their Solutions

Note that the graphs of Equations 1

and 2 are lines.

• As the solution (3, 1)

satisfies each equation,

the point (3, 1) lies on

each line.

• So, it is the point of

intersection of the two

lines.

Note that the graphs of Equations 1

and 2 are lines.

• As the solution (3, 1)

satisfies each equation,

the point (3, 1) lies on

each line.

• So, it is the point of

intersection of the two

lines.

12.
Substitution Method

13.
Substitution Method

To solve a system using the

substitution method, we start with one

equation in the system and solve for one

variable in terms of the other variable.

1. Solve for one variable.

2. Substitute.

3. Back-substitute.

To solve a system using the

substitution method, we start with one

equation in the system and solve for one

variable in terms of the other variable.

1. Solve for one variable.

2. Substitute.

3. Back-substitute.

14.
Step 1

Solve for one variable.

• Choose one equation, and solve

for one variable in terms of the other

variable.

Solve for one variable.

• Choose one equation, and solve

for one variable in terms of the other

variable.

15.
Step 2

• Substitute the expression you found in step 1

into the other equation to get an equation

in one variable.

• Then, solve for that variable.

• Substitute the expression you found in step 1

into the other equation to get an equation

in one variable.

• Then, solve for that variable.

16.
Step 3

• Substitute the value you found in step 2

back into the expression found in step 1

to solve for the remaining variable.

• Substitute the value you found in step 2

back into the expression found in step 1

to solve for the remaining variable.

17.
E.g. 1—Substitution Method

Find all solutions of the system.

2 x y 1 Equation 1

3 x 4 y 14 Equation 2

• We solve for y in the first equation.

y = 1 – 2x

Find all solutions of the system.

2 x y 1 Equation 1

3 x 4 y 14 Equation 2

• We solve for y in the first equation.

y = 1 – 2x

18.
E.g. 1—Substitution Method

Now, we substitute for y in the second

equation and solve for x:

3x + 4(1 – 2x) = 14

3x + 4 – 8x = 14

–5x + 4 = 14

–5x = 10

x = –2

Now, we substitute for y in the second

equation and solve for x:

3x + 4(1 – 2x) = 14

3x + 4 – 8x = 14

–5x + 4 = 14

–5x = 10

x = –2

19.
E.g. 1—Substitution Method

Next, we back-substitute x = –2

into the equation y = 1 – 2x:

y = 1 – 2(–2) = 5

• Thus, x = –2 and y = 5.

• So, the solution is the ordered pair (–2, 5).

Next, we back-substitute x = –2

into the equation y = 1 – 2x:

y = 1 – 2(–2) = 5

• Thus, x = –2 and y = 5.

• So, the solution is the ordered pair (–2, 5).

20.
E.g. 1—Substitution Method

The figure shows that the graphs

of the two equations intersect at

the point (–2, 5).

The figure shows that the graphs

of the two equations intersect at

the point (–2, 5).

21.
Elimination Method

22.
Elimination Method

To solve a system using the elimination

method, we try to combine the equations

using sums or differences so as to eliminate

one of the variables.

1. Adjust the coefficients.

2. Add the equations.

3. Back-substitute.

To solve a system using the elimination

method, we try to combine the equations

using sums or differences so as to eliminate

one of the variables.

1. Adjust the coefficients.

2. Add the equations.

3. Back-substitute.

23.
Step 1

Adjust the coefficients.

• Multiply one or more of the equations by

appropriate numbers so that the coefficient

of one variable in one equation is the negative

of its coefficient in the other equation.

Adjust the coefficients.

• Multiply one or more of the equations by

appropriate numbers so that the coefficient

of one variable in one equation is the negative

of its coefficient in the other equation.

24.
Step 2

Add the equations.

• Add the two equations to eliminate

one variable.

• Then, solve for the remaining variable.

Add the equations.

• Add the two equations to eliminate

one variable.

• Then, solve for the remaining variable.

25.
Step 3

• Substitute the value you found in step 2

back into one of the original equations.

• Then, solve for the remaining variable.

• Substitute the value you found in step 2

back into one of the original equations.

• Then, solve for the remaining variable.

26.
E.g. 2—Elimination Method

Find all solutions of the system.

3 x 2y 14 Equation 1

2 x 4 y 4 Equation 2

• Multiply Equation 2 by ½ so that the coefficients of

the y-terms are negatives of each other.

• Now, we can add the equations to eliminate y.

Find all solutions of the system.

3 x 2y 14 Equation 1

2 x 4 y 4 Equation 2

• Multiply Equation 2 by ½ so that the coefficients of

the y-terms are negatives of each other.

• Now, we can add the equations to eliminate y.

27.
E.g. 2—Elimination Method

3 x 2y 14

x 2y 2

4x 16

x 4

• Now, we back-substitute x = 4 into one of

the original equations and solve for y.

• Let’s choose the second equation because

it looks simpler.

3 x 2y 14

x 2y 2

4x 16

x 4

• Now, we back-substitute x = 4 into one of

the original equations and solve for y.

• Let’s choose the second equation because

it looks simpler.

28.
E.g. 2—Elimination Method

x – 2y = 2

4 – 2y = 2

–2y = –2

y=1

• The solution is (4, 1).

x – 2y = 2

4 – 2y = 2

–2y = –2

y=1

• The solution is (4, 1).

29.
E.g. 2—Elimination Method

The figure shows

that the graphs of

the equations in

the system intersect

at the point (4, 1).

The figure shows

that the graphs of

the equations in

the system intersect

at the point (4, 1).

30.
Graphical Method

31.
Graphical Method

In the graphical method, we use

a graphing device to solve the system

of equations.

In the graphical method, we use

a graphing device to solve the system

of equations.

32.
Graphical Method

1. Graph each equation.

2. Find the intersection points.

1. Graph each equation.

2. Find the intersection points.

33.
Step 1

Graph each equation.

• Express each equation in a form suitable

for the graphing calculator by solving for y

as a function of x.

• Graph the equations on the same screen.

Graph each equation.

• Express each equation in a form suitable

for the graphing calculator by solving for y

as a function of x.

• Graph the equations on the same screen.

34.
Step 2

Find the intersection points.

• The solutions are the x- and y-coordinates

of the points of intersection.

Find the intersection points.

• The solutions are the x- and y-coordinates

of the points of intersection.

35.
E.g. 3—Graphical Method

Find all solutions of the system.

1.35x 2.13 y 2.36

2.16 x 0.32y 1.06

• To graph each equation we solve for y in terms

of x, and we get the equivalent system where

we have rounded the coefficients to

two decimals.

y 0.63 x 1.11

y 6.75 x 3.31

Find all solutions of the system.

1.35x 2.13 y 2.36

2.16 x 0.32y 1.06

• To graph each equation we solve for y in terms

of x, and we get the equivalent system where

we have rounded the coefficients to

two decimals.

y 0.63 x 1.11

y 6.75 x 3.31

36.
E.g. 3—Graphical Method

The figure shows that the two lines

• Zooming in,

we see that

the solution is

approximately

(0.30, 1.30).

The figure shows that the two lines

• Zooming in,

we see that

the solution is

approximately

(0.30, 1.30).

37.
The Number of Solutions of a

Linear System in Two Variables

Linear System in Two Variables

38.
The Number of Solutions of a Linear System in Two Variables

The graph of a linear system in two

variables is a pair of lines.

• So, to solve the system graphically,

we must find the intersection point(s)

of the lines.

The graph of a linear system in two

variables is a pair of lines.

• So, to solve the system graphically,

we must find the intersection point(s)

of the lines.

39.
The Number of Solutions of a Linear System in Two Variables

Two lines may intersect in a single point,

they may be parallel, or they may coincide.

• So, there are three possible outcomes

when solving such a system.

Two lines may intersect in a single point,

they may be parallel, or they may coincide.

• So, there are three possible outcomes

when solving such a system.

40.
The Number of Solutions of a Linear System in Two Variables

For a system of linear equations in two

variables, exactly one of the following is true.

• The system has exactly one solution.

• The system has no solution.

• The system has infinitely many solutions.

For a system of linear equations in two

variables, exactly one of the following is true.

• The system has exactly one solution.

• The system has no solution.

• The system has infinitely many solutions.

41.
Inconsistent and Dependent Systems

A system that has no solution is said

to be inconsistent.

A system with infinitely many solutions

is called dependent.

A system that has no solution is said

to be inconsistent.

A system with infinitely many solutions

is called dependent.

42.
E.g. 4—A Linear System with One Solution

Solve the system and graph

the lines.

3 x y 0 Equation 1

5 x 2y 22 Equation 2

Solve the system and graph

the lines.

3 x y 0 Equation 1

5 x 2y 22 Equation 2

43.
E.g. 4—A Linear System with One Solution

We eliminate y from the equations and

solve for x.

6 x 2y 0 2 Equation1

5 x 2 y 22

11x 22

x 2

We eliminate y from the equations and

solve for x.

6 x 2y 0 2 Equation1

5 x 2 y 22

11x 22

x 2

44.
E.g. 4—A Linear System with One Solution

Now, we back-substitute into the first

equation and solve for y:

6(2) – 2y = 0

–2y = –12

y =6

• The solution of the system is the ordered pair

(2, 6).

• That is, x = 2, y = 6

Now, we back-substitute into the first

equation and solve for y:

6(2) – 2y = 0

–2y = –12

y =6

• The solution of the system is the ordered pair

(2, 6).

• That is, x = 2, y = 6

45.
E.g. 4—A Linear System with One Solution

The graph shows that

the lines in the system

intersect at the point

(2, 6).

The graph shows that

the lines in the system

intersect at the point

(2, 6).

46.
E.g. 5—A Linear System with No Solution

Solve the system.

8 x 2y 5 Equation 1

12 x 3 y 7 Equation 2

• This time, we try to find a suitable combination

of the two equations to eliminate the variable y.

Solve the system.

8 x 2y 5 Equation 1

12 x 3 y 7 Equation 2

• This time, we try to find a suitable combination

of the two equations to eliminate the variable y.

47.
E.g. 5—A Linear System with No Solution

Multiplying the first equation by 3 and

the second by 2 gives:

24 x 6 y 15

24 x 6 y 14

0 29

• Adding the two equations eliminates

both x and y in this case.

• So, we end up with 0 = 29, which is obviously

false.

Multiplying the first equation by 3 and

the second by 2 gives:

24 x 6 y 15

24 x 6 y 14

0 29

• Adding the two equations eliminates

both x and y in this case.

• So, we end up with 0 = 29, which is obviously

false.

48.
E.g. 5—A Linear System with No Solution

No matter what values we assign to x

and y, we cannot make this statement

• So, the system has no solution.

No matter what values we assign to x

and y, we cannot make this statement

• So, the system has no solution.

49.
E.g. 5—A Linear System with No Solution

The figure shows that the lines in

the system are parallel and do not

• The system is

inconsistent.

The figure shows that the lines in

the system are parallel and do not

• The system is

inconsistent.

50.
E.g. 6—A Linear System with Infinitely Many Solutions

Solve the system.

3 x 6 y 12 Equation 1

4 x 8 y 16 Equation 2

• We multiply the first equation by 4 and

the second by 3 to prepare for subtracting

the equations to eliminate x.

Solve the system.

3 x 6 y 12 Equation 1

4 x 8 y 16 Equation 2

• We multiply the first equation by 4 and

the second by 3 to prepare for subtracting

the equations to eliminate x.

51.
E.g. 6—A Linear System with Infinitely Many Solutions

The new equations are:

12 x 24 y 48

12 x 24 y 48

• We see that the two equations in the original

system are simply different ways of expressing

the equation of one single line.

• The coordinates of any point on this line give

a solution of the system.

The new equations are:

12 x 24 y 48

12 x 24 y 48

• We see that the two equations in the original

system are simply different ways of expressing

the equation of one single line.

• The coordinates of any point on this line give

a solution of the system.

52.
E.g. 6—A Linear System with Infinitely Many Solutions

Writing the equation in slope-intercept

form, we have: y = ½x – 2

• So, if we let t represent any real number,

we can write the solution as:

x=t

y = ½t – 2

• We can also write the solution

in ordered-pair form as:

(t, ½t – 2)

where t is any real number.

Writing the equation in slope-intercept

form, we have: y = ½x – 2

• So, if we let t represent any real number,

we can write the solution as:

x=t

y = ½t – 2

• We can also write the solution

in ordered-pair form as:

(t, ½t – 2)

where t is any real number.

53.
E.g. 6—A Linear System with Infinitely Many Solutions

The system has infinitely many

The system has infinitely many

54.
A Linear System with Infinitely Many

Solutions

In Example 3, to get specific solutions,

we have to assign values to t.

• If t = 1, we get

3

the solution (1, – 2 ).

• If t = 4, we get

the solution (4, 0).

• For every value of t,

we get a different

solution.

Solutions

In Example 3, to get specific solutions,

we have to assign values to t.

• If t = 1, we get

3

the solution (1, – 2 ).

• If t = 4, we get

the solution (4, 0).

• For every value of t,

we get a different

solution.

55.
Modeling with Linear Systems

56.
Modeling with Linear Systems

Frequently, when we use equations to

solve problems in the sciences or in other

areas, we obtain systems like the ones

we’ve been considering.

Frequently, when we use equations to

solve problems in the sciences or in other

areas, we obtain systems like the ones

we’ve been considering.

57.
Guidelines for Modeling with Systems of Equations

When modeling with systems of equations,

we use these guidelines—similar to

those in Section 1.5.

1. Identify the variables.

2. Express all unknown quantities in terms of

the variables.

3. Set up a system of equations.

4. Solve the system and interpret the results.

When modeling with systems of equations,

we use these guidelines—similar to

those in Section 1.5.

1. Identify the variables.

2. Express all unknown quantities in terms of

the variables.

3. Set up a system of equations.

4. Solve the system and interpret the results.

58.
Guideline 1 for Modeling with Systems of Equations

Identify the variables.

• Identify the quantities the problem asks you to find.

• These are usually determined by a careful reading

of the question posed at the end of the problem.

• Introduce notation for the variables.

(Call them x and y or some other letters).

Identify the variables.

• Identify the quantities the problem asks you to find.

• These are usually determined by a careful reading

of the question posed at the end of the problem.

• Introduce notation for the variables.

(Call them x and y or some other letters).

59.
Guideline 2 for Modeling with Systems of Equations

Express all unknown quantities

in terms of the variables.

• Read the problem again, and express all

the quantities mentioned in the problem

in terms of the variables you defined in step 1.

Express all unknown quantities

in terms of the variables.

• Read the problem again, and express all

the quantities mentioned in the problem

in terms of the variables you defined in step 1.

60.
Guideline 3 for Modeling with Systems of Equations

Set up a system of equations.

• Find the crucial facts in the problem that give

the relationships between the expressions

you found in Step 2.

• Set up a system of equations (or a model)

that expresses these relationships.

Set up a system of equations.

• Find the crucial facts in the problem that give

the relationships between the expressions

you found in Step 2.

• Set up a system of equations (or a model)

that expresses these relationships.

61.
Guideline 4 for Modeling with Systems of Equations

Solve the system and interpret

the results.

• Solve the system you found in Step 3.

• Check your solutions.

• State your final answer as a sentence that

answers the question posed in the problem.

Solve the system and interpret

the results.

• Solve the system you found in Step 3.

• Check your solutions.

• State your final answer as a sentence that

answers the question posed in the problem.

62.
E.g. 7—A Distance-Speed-Time Problem

A woman rows a boat upstream from

one point on a river to another point 4 mi

away in 1½ hours.

• The return trip,

traveling with

the current,

takes only

45 min.

A woman rows a boat upstream from

one point on a river to another point 4 mi

away in 1½ hours.

• The return trip,

traveling with

the current,

takes only

45 min.

63.
E.g. 7—A Distance-Speed-Time Problem

How fast does she row relative to the water?

At what speed is the current flowing?

How fast does she row relative to the water?

At what speed is the current flowing?

64.
E.g. 7—A Distance-Speed-Time Problem

We are asked to find the rowing speed

and the speed of the current.

• So, we let:

x = rowing speed (mi/h)

y = current speed (mi/h)

We are asked to find the rowing speed

and the speed of the current.

• So, we let:

x = rowing speed (mi/h)

y = current speed (mi/h)

65.
E.g. 7—A Distance-Speed-Time Problem

The woman’s speed when she rows

upstream is:

• Her rowing speed minus the speed of the current

Her speed downstream is:

• Her rowing speed plus the speed of the current.

The woman’s speed when she rows

upstream is:

• Her rowing speed minus the speed of the current

Her speed downstream is:

• Her rowing speed plus the speed of the current.

66.
E.g. 7—A Distance-Speed-Time Problem

Now, we translate this information into

the language of algebra.

Now, we translate this information into

the language of algebra.

67.
E.g. 7—A Distance-Speed-Time Problem

The distance upstream and downstream

is 4 mi.

• So, using the fact that

speed time = distance for both legs of the trip

we get:

speed upstream time upstream distance traveled

speed downstream time downstream distance traveled

The distance upstream and downstream

is 4 mi.

• So, using the fact that

speed time = distance for both legs of the trip

we get:

speed upstream time upstream distance traveled

speed downstream time downstream distance traveled

68.
E.g. 7—A Distance-Speed-Time Problem

In algebraic notation, that translates into

these equations.

( x y ) 32 4 Equation 1

( x y ) 34 4 Equation 2

• The times have been converted to hours, since

we are expressing the speeds in miles per hour.

In algebraic notation, that translates into

these equations.

( x y ) 32 4 Equation 1

( x y ) 34 4 Equation 2

• The times have been converted to hours, since

we are expressing the speeds in miles per hour.

69.
E.g. 7—A Distance-Speed-Time Problem

We multiply the equations by 2 and 4,

respectively, to clear the denominators.

3 x 3 y 8

3 x 3 y 16

6x 24

x 4

We multiply the equations by 2 and 4,

respectively, to clear the denominators.

3 x 3 y 8

3 x 3 y 16

6x 24

x 4

70.
E.g. 7—A Distance-Speed-Time Problem

Back-substituting this value of x into

the first equation (the second works

just as well) and solving for y gives:

3(4) 3 y 8

3 y 8 12

y 34

• The woman rows at 4 mi/h.

• The current flows at 1⅓ mi/h.

Back-substituting this value of x into

the first equation (the second works

just as well) and solving for y gives:

3(4) 3 y 8

3 y 8 12

y 34

• The woman rows at 4 mi/h.

• The current flows at 1⅓ mi/h.

71.
E.g. 8—A Mixture Problem

A vintner fortifies wine that contains 10%

alcohol by adding 70% alcohol solution to it.

The resulting mixture has an alcoholic

strength of 16% and fills 1000 one-liter

• How many liters (L) of the wine and

of the alcohol solution does vintner use?

A vintner fortifies wine that contains 10%

alcohol by adding 70% alcohol solution to it.

The resulting mixture has an alcoholic

strength of 16% and fills 1000 one-liter

• How many liters (L) of the wine and

of the alcohol solution does vintner use?

72.
E.g. 8—A Mixture Problem

Since we are asked for the amounts

of wine and alcohol, we let:

x = amount of wine used (L)

y = amount of alcohol solution used (L)

Since we are asked for the amounts

of wine and alcohol, we let:

x = amount of wine used (L)

y = amount of alcohol solution used (L)

73.
E.g. 8—A Mixture Problem

From the fact that the wine contains 10%

alcohol and the solution contains 70%

alcohol, we get the following.

From the fact that the wine contains 10%

alcohol and the solution contains 70%

alcohol, we get the following.

74.
E.g. 8—A Mixture Problem

The volume of the mixture must be

the total of the two volumes the vintner

is adding together.

• Thus,

x + y = 1000

The volume of the mixture must be

the total of the two volumes the vintner

is adding together.

• Thus,

x + y = 1000

75.
E.g. 8—A Mixture Problem

Also, the amount of alcohol in the mixture

must be the total of the alcohol contributed

by the wine and by the alcohol solution.

• That is,

0.10x + 0.70y = (0.16)1000

0.10x + 0.70y = 160

x + 7y = 1600

Also, the amount of alcohol in the mixture

must be the total of the alcohol contributed

by the wine and by the alcohol solution.

• That is,

0.10x + 0.70y = (0.16)1000

0.10x + 0.70y = 160

x + 7y = 1600

76.
E.g. 8—A Mixture Problem

Thus, we get the system

x y 1000 Equation1

x 7 y 1600 Equation 2

• Subtracting the first equation from the second

eliminates the variable x, and we get:

6y = 600

y = 100

Thus, we get the system

x y 1000 Equation1

x 7 y 1600 Equation 2

• Subtracting the first equation from the second

eliminates the variable x, and we get:

6y = 600

y = 100

77.
E.g. 8—A Mixture Problem

We now back-substitute y = 100 into

the first equation and solve for x:

x + 100 = 1000

x = 900

• The vintner uses 900 L of wine

and 100 L of the alcohol solution.

We now back-substitute y = 100 into

the first equation and solve for x:

x + 100 = 1000

x = 900

• The vintner uses 900 L of wine

and 100 L of the alcohol solution.