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5: Systems of Two Linear Equations

5: Systems of Two Linear Equations

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8.5 • Systems of Two Linear Equations



351



To check this, we can substitute 6 for x and 1 for y in both equations.

x Ϫ 2y ϭ 4 becomes 6 Ϫ 2(1) ϭ 4

x ϩ 2y ϭ 8 becomes 6 ϩ 2(1) ϭ 8



A true statement

A true statement



y

x + 2y = 8

(6, 1)

x



x − 2y = 4



Figure 8.45



Thus we say that {(6, 1)} is the solution set of the system

a



x Ϫ 2y ϭ 4

b

x ϩ 2y ϭ 8



Two or more linear equations in two variables considered together are called a system of

linear equations. Here are three systems of linear equations:

a



x Ϫ 2y ϭ 4

b

x ϩ 2y ϭ 8



a



4x Ϫ y ϭ 5

° 2x ϩ y ϭ 9 ¢

7x Ϫ 2y ϭ 13



5x Ϫ 3y ϭ 9

b

3x ϩ 7y ϭ 12



To solve a system of linear equations means to find all of the ordered pairs that are solutions of all of the equations in the system. There are several techniques for solving systems

of linear equations. We will use three of them in this chapter: two methods are presented in

this section, and a third method is presented in Section 8.6.

To solve a system of linear equations by graphing, we proceed as in the opening discussion of this section. We graph the equations on the same set of axes, and then the ordered pairs

associated with any points of intersection are the solutions to the system. Let’s consider

another example.



Classroom Example

x ϩ 2y ϭ 5

Solve the system a

b.

x Ϫ y ϭ Ϫ4



EXAMPLE 1



Solve the system a



xϩ yϭ 5

b.

x Ϫ 2y ϭ Ϫ4



Solution

We can find the intercepts and a check point for each of the lines.

x ؊ 2y ‫ ؍‬؊4



x؉y‫؍‬5

x



y



0

5

2



5

0

3



Intercepts

Check point



x



y



0

Ϫ4

Ϫ2



2

0

1



Intercepts

Check point



352



Chapter 8 • Coordinate Geometry and Linear Systems



Figure 8.46 shows the graphs of the two equations. It appears that (2, 3) is the solution of the

system. To check it we can substitute 2 for x and 3 for y in both equations.

y

x+y=5



x − 2y = − 4



(2, 3)

x



Figure 8.46



xϩy ϭ5

becomes 2 ϩ 3 ϭ 5

x Ϫ 2y ϭ Ϫ4 becomes 2 Ϫ 2(3) ϭ Ϫ4



A true statement

A true statement



Therefore, {(2, 3)} is the solution set.

It should be evident that solving systems of equations by graphing requires accurate

graphs. In fact, unless the solutions are integers, it is really quite difficult to obtain exact solutions from a graph. For this reason the systems to solve by graphing in this section have integer solutions. Furthermore, checking a solution takes on additional significance when the

graphing approach is used. By checking, you can be absolutely sure that you are reading the

correct solution from the graph.

Figure 8.47 shows the three possible cases for the graph of a system of two linear equations in two variables.

Case I



The graphs of the two equations are two lines intersecting in one point. There is

one solution, and we call the system a consistent system.



Case II



The graphs of the two equations are parallel lines. There is no solution, and we call

the system an inconsistent system.



Case III



The graphs of the two equations are the same line. There are infinitely many solutions to the system. Any pair of real numbers that satisfies one of the equations also

satisfies the other equation, and we say the equations are dependent.



y



y



x



Case I

Figure 8.47



y



x



Case II



x



Case III



8.5 • Systems of Two Linear Equations



353



Thus as we solve a system of two linear equations in two variables, we know what to expect.

The system will have no solutions, one ordered pair as a solution, or infinitely many ordered

pairs as solutions. Most of the systems that we will be working with in this text have one

solution.

An example of case I was given in Example 1 (Figure 8.46). The next two examples

illustrate the other cases.



Classroom Example

5x Ϫ y ϭ 2

Solve the system a

b.

5x Ϫ y ϭ 5



Solve the system a



EXAMPLE 2



2x ϩ 3y ϭ 6

b.

2x ϩ 3y ϭ 12



Solution

2x ؉ 3y ‫ ؍‬6



2x ؉ 3y ‫ ؍‬12



x



y



x



y



0

3

Ϫ3



2

0

4



0

6

3



4

0

2



Figure 8.48 shows the graph of the system. Since the lines are parallel, there is no solution to

the system. The solution set is л.

y



2x + 3y = 12



x

2x + 3y = 6

Figure 8.48



Classroom Example

3x Ϫ 2y ϭ 4

Solve the system a

b.

9x Ϫ 6y ϭ 12



EXAMPLE 3



Solve the system a



xϩ yϭ3

b.

2x ϩ 2y ϭ 6



Solution

x؉y‫؍‬3



2x ؉ 2y ‫ ؍‬6



x



y



x



y



0

3

1



3

0

2



0

3

1



3

0

2



Figure 8.49 shows the graph of this system. Since the graphs of both equations are the same

line, there are infinitely many solutions to the system. Any ordered pair of real numbers that

satisfies one equation also satisfies the other equation.



354



Chapter 8 • Coordinate Geometry and Linear Systems



y



x+y=3



x

2 x + 2y = 6



Figure 8.49



Substitution Method

As stated earlier, solving systems of equations by graphing requires accurate graphs. In fact,

unless the solutions are integers, it is quite difficult to obtain exact solutions from a graph.

Thus we will consider some other methods for solving systems of equations.

The substitution method works quite well with systems of two linear equations in two

unknowns.

Step 1 Solve one of the equations for one variable in terms of the other variable, if neither

equation is in such a form. (If possible, make a choice that will avoid fractions.)

Step 2 Substitute the expression obtained in step 1 into the other equation. This produces an

equation in one variable.

Step 3 Solve the equation obtained in step 2.

Step 4 Use the solution obtained in step 3, along with the expression obtained in step 1, to

determine the solution of the system.

Now let’s look at some examples that illustrate the substitution method.

Classroom Example

3x ϩ y ϭ 4

Solve the system a

b by

yϭx Ϫ8

the substitution method.



EXAMPLE 4



Solve the system a



x ϩ y ϭ 16

b.

y ϭ x ϩ2



Solution

Because the second equation states that y equals x ϩ 2, we can substitute x ϩ 2 for y in the

first equation.

x ϩ y ϭ 16



Substitute x ϩ 2 for y



x ϩ (x ϩ 2) ϭ 16



Now we have an equation with one variable that we can solve in the usual way.

x ϩ (x ϩ 2) ϭ 16

2x ϩ 2 ϭ 16

2x ϭ 14

xϭ7

Substituting 7 for x in one of the two original equations (let’s use the second one) yields

yϭ7ϩ2ϭ9



✔ Check

To check, we can substitute 7 for x and 9 for y in both of the original equations.

7 ϩ 9 ϭ 16 A true statement

9 ϭ 7 ϩ 2 A true statement

The solution set is {(7, 9)}.



8.5 • Systems of Two Linear Equations



Classroom Example

2x Ϫ 9y ϭ 14

Solve the system a

b

3x Ϫ y ϭ 6

by the substitution method.



Solve the system a



EXAMPLE 5

Solution



3x Ϫ 7y ϭ 2

b.

x ϩ 4y ϭ 1



Let’s solve the second equation for x in terms of y.

x ϩ 4y ϭ 1

x ϭ 1 Ϫ 4y

Now we can substitute 1 Ϫ 4y for x in the first equation.

3x Ϫ 7y ϭ 2



Substitute 1 Ϫ 4y for x



3(1 Ϫ 4y) Ϫ 7y ϭ 2



Let’s solve this equation for y.

3(1 Ϫ 4y) Ϫ 7y ϭ 2

3 Ϫ 12y Ϫ 7y ϭ 2

Ϫ19y ϭ Ϫ1

1



19

Finally, we can substitute

x ϭ 1 Ϫ 4a

xϭ1Ϫ





1

b

19



4

19



15

19



The solution set is ea



Classroom Example

8x Ϫ 3y ϭ 14

Solve the system a

b

2x ϩ 5y ϭ Ϫ8

by the substitution method.



1

for y in the equation x ϭ 1 Ϫ 4y.

19



EXAMPLE 6



15 1

, bf .

19 19



Solve the system a



Solution



5x Ϫ 6y ϭ Ϫ 4

b.

3x ϩ 2y ϭ Ϫ 8



Note that solving either equation for either variable will produce a fractional form.

Let’s solve the second equation for y in terms of x.

3x ϩ 2y ϭ Ϫ8

2y ϭ Ϫ8 Ϫ 3x

Ϫ8 Ϫ 3x



2

Ϫ8 Ϫ 3x

Now we can substitute

for y in the first equation.

2

Substitute



5x Ϫ 6y ϭ Ϫ4



Ϫ8 Ϫ 3x

for y

2



Solving the equation yields

Ϫ8 Ϫ 3x

b ϭ Ϫ4

2

5x Ϫ 3(Ϫ8 Ϫ 3x) ϭ Ϫ4

5x ϩ 24 ϩ 9x ϭ Ϫ4

14x ϭ Ϫ28

x ϭ Ϫ2



5x Ϫ 6 a



5x Ϫ 6 a



Ϫ8 Ϫ 3x

b ϭ Ϫ4

2



355



356



Chapter 8 • Coordinate Geometry and Linear Systems



Substituting Ϫ2 for x in y ϭ



Ϫ8 Ϫ 3x

yields

2



Ϫ8 Ϫ 3(Ϫ2)

2

Ϫ8 ϩ 6



2

Ϫ2



2

y ϭ Ϫ1





The solution set is {(Ϫ2, Ϫ1)}.

Classroom Example

x ϩ 3y ϭ 9

Solve the system a

b

2x ϩ 6y ϭ 11

by the substitution method.



Solve the system a



EXAMPLE 7



2x ϩ y ϭ 4

b.

4x ϩ 2y ϭ 7



Solution

Let’s solve the first equation for y in terms of x.

2x ϩ y ϭ 4

y ϭ 4 Ϫ 2x

Now we can substitute 4 Ϫ 2x for y in the second equation.

4x ϩ 2y ϭ 7



Substitute 4 Ϫ 2x for y



4x ϩ 2(4 Ϫ 2x) ϭ 7



Let’s solve this equation for x.

4x ϩ 2(4 Ϫ 2x) ϭ 7

4x ϩ 8 Ϫ 4x ϭ 7

8ϭ7

The statement 8 ϭ 7 is a contradiction, and therefore the original system is inconsistent; it

has no solutions. The solution set is Ø.

Classroom Example

8x Ϫ 2y ϭ 6

Solve the system a

b

4x ϭ y ϩ 3

by the substitution method.



Solve the system a



EXAMPLE 8



y ϭ 2x ϩ 1

b.

4x Ϫ 2y ϭ Ϫ 2



Solution

Because the first equation states that y equals 2x ϩ 1, we can substitute 2x ϩ 1 for y in the

second equation.

4x Ϫ 2y ϭ Ϫ2



Substitute 2x ϩ 1 for y



4x Ϫ 2(2x ϩ 1) ϭ Ϫ2



Let’s solve this equation for x.

4x Ϫ 2(2x ϩ 1) ϭ Ϫ2

4x Ϫ 4x Ϫ 2 ϭ Ϫ2

Ϫ2 ϭ Ϫ2

We obtained a true statement, Ϫ2 ϭ Ϫ2, which indicates that the system has an infinite number of solutions. Any ordered pair that satisfies one of the equations will also satisfy the other

equation. Thus the solution set is any ordered pair on the line y ϭ 2x ϩ 1, and the solution

set can be written as {(x, y)Ϳ y ϭ 2x ϩ 1}.



Problem Solving

Many word problems that we solved earlier in this text by using one variable and one equation can also be solved by using a system of two linear equations in two variables. In fact, in

many of these problems you may find it much more natural to use two variables. Let’s consider some examples.



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8.5 • Systems of Two Linear Equations



Classroom Example

Sonora invested some money at 5%

and $600 less than that amount at

3%. The yearly interest from the two

investments was $190. How much

did Sonora invest at each rate?



357



EXAMPLE 9

Anita invested some money at 8% and $400 more than that amount at 9%. The yearly interest from the two investments was $87. How much did Anita invest at each rate?



Solution

Let x represent the amount invested at 8%, and let y represent the amount invested at 9%. The

problem translates into this system:

y ϭ x ϩ 400

The amount invested at 9% was $400 more than at 8%

a

b

The yearly interest from the two investments was $87

0.08x ϩ 0.09y ϭ 87

From the first equation, we can substitute x ϩ 400 for y in the second equation and solve for x.

0.08x ϩ 0.09(x ϩ 400) ϭ 87

0.08x ϩ 0.09x ϩ 36 ϭ 87

0.17x ϭ 51

x ϭ 300

Therefore, Anita invested $300 at 8% and $300 ϩ $400 ϭ $700 at 9%.



Classroom Example

The proceeds of ticket sales at the

children’s theater were $6296. The

price of a child’s ticket was $12 and

adult tickets were $20. If a total of

430 children’s and adult tickets were

sold, how many of each were sold?



EXAMPLE 10

The proceeds from a concession stand that sold hamburgers and hot dogs at the baseball game

were $575.50. The price of a hot dog was $2.50, and the price of a hamburger was $3.00. If

a total of 213 hot dogs and hamburgers were sold, how many of each kind were sold?



Solution

Let x equal the number of hot dogs sold, and let y equal the number of hamburgers sold. The

problem translates into this system:

The number sold

The proceeds from the sales



a



x ϩ y ϭ 213

b

2.50x ϩ 3.00y ϭ 575.50



Let’s begin by solving the first equation for y.

x ϩ y ϭ 213

y ϭ 213 Ϫ x

Now we will substitute 213 Ϫ x for y in the second equation and solve for x.

2.50x ϩ 3.00(213 Ϫ x) ϭ 575.50

2.50x ϩ 639.00 Ϫ 3.00x ϭ 575.50

Ϫ0.5x ϩ 639.00 ϭ 575.50

Ϫ0.5x ϭ Ϫ63.50

x ϭ 127

Therefore, there were 127 hot dogs sold and 213 Ϫ 127 ϭ 86 hamburgers sold.



Concept Quiz 8.5

For Problems 1–10, answer true or false.

1. To solve a system of equations means to find all the ordered pairs that satisfy all of the

equations in the system.

2. A consistent system of linear equations will have more than one solution.

3. If the graph of a system of two linear equations results in two distinct parallel lines,

then the system has no solution.

4. Every system of equations has a solution.



358



Chapter 8 • Coordinate Geometry and Linear Systems



5. If the graphs of the two equations in a system are the same line, then the equations in

the system are dependent.

6. To solve a system of two equations in variables x and y, it is sufficient to just find a value for x.

7. For the system a



2x ϩ y ϭ 4

b , the ordered pair (1, 2) is a solution.

x ϩ 5y ϭ 10



8. Graphing a system of equations is the most accurate method to find the solution of the

system.

9. The solution set of the system a

10. The system a



x ϩ 2y ϭ 4

b is the null set.

2x ϩ 4y ϭ 9



2x ϩ 2y ϭ Ϫ4

b has infinitely many solutions.

x ϭ Ϫy Ϫ2



Problem Set 8.5

For Problems 1–20, use the graphing method to solve each

system. (Objective 1)

xϩyϭ1

xϪyϭ 2

b

b

1. a

2. a

xϪyϭ3

x ϩ y ϭ Ϫ4

3. a



x ϩ 2y ϭ 4

b

2x Ϫ y ϭ 3



4. a



2x Ϫ y ϭ Ϫ8

b

xϩyϭ 2



5. a



x ϩ 3y ϭ 6

b

x ϩ 3y ϭ 3



6. a



y ϭ Ϫ2x

b

y Ϫ 3x ϭ 0



7. a



xϩyϭ0

b

xϪyϭ0



8. a



3x Ϫ y ϭ 3

b

3x Ϫ y ϭ Ϫ3



3x Ϫ 2y ϭ 5

b

9. a

2x ϩ 5y ϭ Ϫ3



2x ϩ 3y ϭ 1

b

10. a

4x Ϫ 3y ϭ Ϫ7



11. a



y ϭ Ϫ2x ϩ 3

b

6x ϩ 3y ϭ 9



12. a



y ϭ 2x ϩ 5

b

x ϩ 3y ϭ Ϫ6



13. a



y ϭ 5x Ϫ 2

b

4x ϩ 3y ϭ 13



14. a



yϭxϪ2

b

2x Ϫ 2y ϭ 4



y ϭ 4 Ϫ 2x

b

15. a

y ϭ 7 Ϫ 3x



y ϭ 3x ϩ 4

b

16. a

y ϭ 5x ϩ 8



25. a



x ϭ Ϫ3y

b

7x Ϫ 2y ϭ Ϫ69



26. a



9x Ϫ 2y ϭ Ϫ38

b

y ϭ Ϫ5x



27. a



x ϩ 2y ϭ5

b

3x ϩ 6y ϭ Ϫ2



28. a



4x ϩ 2y ϭ 6

b

y ϭ Ϫ2x ϩ 3



29. a



3x Ϫ 4y ϭ 9

b

x ϭ 4y Ϫ 1



30. a



y ϭ 3x Ϫ 5

b

2x ϩ 3y ϭ 6



2

yϭ xϪ1

5

¢

31. °

3x ϩ 5y ϭ 4

3

yϭ x Ϫ 5

4

¢

32. °

5x Ϫ 4y ϭ 9

33. °



7x Ϫ 3y ϭ Ϫ2

¢

3

xϭ yϩ1

4



34. °



5x Ϫ y ϭ 9

¢

1

x ϭ yϪ3

2



17. a



y ϭ 2x

b

3x Ϫ 2y ϭ Ϫ2



18. a



y ϭ 3x

b

4x Ϫ 3y ϭ 5



35. a



2x ϩ y ϭ 12

b

3x Ϫ y ϭ 13



19. a



7x Ϫ 2y ϭ Ϫ8

b

x ϭ Ϫ2



20. a



3x ϩ 8y ϭ Ϫ1

b

y ϭ Ϫ2



36. a



Ϫx ϩ 4y ϭ Ϫ22

b

x Ϫ 7y ϭ 34



37. a



4x ϩ 3y ϭ Ϫ40

b

5x Ϫ y ϭ Ϫ12



For Problems 21– 46, solve each system by using the substitution method. (Objective 2)

21. a



x ϩ y ϭ 20

b

xϭyϪ4



22. a



x ϩ y ϭ 23

b

yϭxϪ5



38. a



x Ϫ 5y ϭ 33

b

Ϫ4x ϩ 7y ϭ Ϫ41



23. a



y ϭ Ϫ3x Ϫ18

b

5x Ϫ 2y ϭ Ϫ8



24. a



4x Ϫ 3y ϭ 33

b

x ϭ Ϫ4y Ϫ 25



39. a



3x ϩ y ϭ 2

b

11x Ϫ 3y ϭ 5



8.5 • Systems of Two Linear Equations



40. a



2x Ϫ y ϭ 9

b

7x ϩ 4y ϭ 1



41. a



4x Ϫ 8y ϭ Ϫ12

b

3x Ϫ 6y ϭ Ϫ9



42. a



2x Ϫ 4y ϭ Ϫ6

b

3x Ϫ 6y ϭ 10



50. The difference of two numbers is 75. The larger number

is 3 less than four times the smaller number. Find the

numbers.

51. In a class of 50 students, the number of women is 2

more than five times the number of men. How many

women are there in the class?

52. In a recent survey, 1000 registered voters were asked

about their political preferences. The number of men in

the survey was 5 less than one-half the number of

women. Find the number of men in the survey.



4x Ϫ 5y ϭ 3

43. a

b

8x ϩ 15y ϭ Ϫ24

44. a



2x ϩ 3y ϭ 3

b

4x Ϫ 9y ϭ Ϫ4



45. a



6x Ϫ 3y ϭ 4

b

5x ϩ 2y ϭ Ϫ1



359



53. The perimeter of a rectangle is 94 inches. The length of

the rectangle is 7 inches more than the width. Find the

dimensions of the rectangle.

54. Two angles are supplementary, and the measure of one

of them is 20o less than three times the measure of the

other angle. Find the measure of each angle.



7x Ϫ 2y ϭ 1

46. a

b

4x ϩ 5y ϭ 2

For Problems 47–58, solve each problem by setting up and

solving an appropriate system of linear equations.

(Objective 5)



47. Doris invested some money at 7% and some money at

8%. She invested $6000 more at 8% than she did at 7%.

Her total yearly interest from the two investments was

$780. How much did Doris invest at each rate?

48. Suppose that Gus invested a total of $8000, part of it at

4% and the remainder at 6%. His yearly income from

the two investments was $380. How much did he invest

at each rate?

49. Find two numbers whose sum is 131 such that one number is 5 less than three times the other.



55. A deposit slip listed $700 in cash to be deposited. There

were 100 bills, some of them were five-dollar bills and

the remainder were ten-dollar bills. How many bills of

each denomination were deposited?

56. Cindy has 30 coins, consisting of dimes and quarters,

which total $5.10. How many coins of each kind does

she have?

57. The income from a student production was $27,500.

The price of a student ticket was $8, and nonstudent

tickets were sold at $15 each. Three thousand tickets

were sold. How many tickets of each kind were sold?

58. Sue bought 3 packages of cookies and 2 sacks of potato chips for $7.35. Later she bought 2 packages of cookies and 5 sacks of potato chips for $9.63. Find the price

of a package of cookies.



Thoughts Into Words

59. Discuss the strengths and weaknesses of solving a system of linear equations by graphing.

60. Determine a system of two linear equations for which

the solution set is {(5, 7)}. Are there other systems that

have the same solution set? If so, find at least one more

system.

61. Give a general description of how to use the substitution

method to solve a system of two linear equations in two

variables.



Answers to the Concept Quiz

1. True

2. False

3. True

4. False

9. True

10. True



5. True



62. Is it possible for a system of two linear equations in two

variables to have exactly two solutions? Defend your

answer.

63. Explain how you would use the substitution method to

solve the system

a



2x ϩ 5y ϭ 5

b

5x Ϫ y ϭ 9



6. False



7. False



8. False



360



Chapter 8 • Coordinate Geometry and Linear Systems



8.6



Elimination-by-Addition Method



OBJECTIVES



1



Solve linear systems of equations by the elimination-by-addition method



2



Solve word problems using a system of two linear equations



We found in the previous section that the substitution method for solving a system of two

equations and two unknowns works rather well. However, as the number of equations and

unknowns increases, the substitution method becomes quite unwieldy. In this section we are

going to introduce another method, called the elimination-by-addition method. We shall

introduce it here, using systems of two linear equations in two unknowns. Later in the text,

we shall extend its use to three linear equations in three unknowns.

The elimination-by-addition method involves replacing systems of equations with simpler

equivalent systems until we obtain a system from which we can easily extract the solutions.

Equivalent systems of equations are systems that have exactly the same solution set. We

can apply the following operations or transformations to a system of equations to produce an

equivalent system.

1. Any two equations of the system can be interchanged.

2. Both sides of any equation of the system can be multiplied by any nonzero real number.

3. Any equation of the system can be replaced by the sum of the equation and a nonzero multiple of another equation.

Now let’s see how to apply these operations to solve a system of two linear equations in two

unknowns.

Classroom Example

6x ϩ 9y ϭ 9

Solve the system a

b

4x Ϫ 9y ϭ 21

by the elimination-by-addition

method.



EXAMPLE 1



Solve the system a



3x ϩ 2y ϭ 1

b.

5x Ϫ 2y ϭ 23



(1)

(2)



Solution

Let’s replace equation (2) with an equation we form by multiplying equation (1) by 1 and

then adding that result to equation (2).

a



3x ϩ 2y ϭ 1

b

8x

ϭ 24



(3)

(4)



From equation (4) we can easily obtain the value of x.

8x ϭ 24

xϭ3

Then we can substitute 3 for x in equation (3).

3x ϩ 2y ϭ 1

3(3) ϩ 2y ϭ 1

2y ϭ Ϫ8

y ϭ Ϫ4

The solution set is {(3, Ϫ4)}. Check it!

Classroom Example

2x Ϫ 5y ϭ 28

Solve the system a

b

8x ϩ y ϭ 28

using the elimination-by-addition

method.



EXAMPLE 2



Solve the system a



x ϩ 5y ϭ Ϫ2

b.

3x Ϫ 4y ϭ Ϫ25



(1)

(2)



Solution

Let’s replace equation (2) with an equation we form by multiplying equation (1) by Ϫ3 and

then adding that result to equation (2).



8.6 • Elimination-by-Addition Method



a



x ϩ 5y ϭ Ϫ2

b

Ϫ19y ϭ Ϫ19



361



(3)

(4)



From equation (4) we can obtain the value y.

Ϫ19y ϭ Ϫ19

yϭ1

Now we can substitute 1 for y in equation (3).

x ϩ 5y ϭ Ϫ2

x ϩ 5(1) ϭ Ϫ2

x ϭ Ϫ7

The solution set is {(Ϫ7, 1)}.



Note that our objective has been to produce an equivalent system of equations such that

one of the variables can be eliminated from one equation. We accomplish this by multiplying

one equation of the system by an appropriate number and then adding that result to the other

equation. Thus the method is called elimination by addition. Let’s look at another example.



Classroom Example

3x Ϫ 5y ϭ 31

Solve the system a

b

7x ϩ 2y ϭ 4

using the elimination-by-addition

method.



EXAMPLE 3



Solve the system a



2x ϩ 5y ϭ 4

b.

5x Ϫ 7y ϭ Ϫ29



(1)

(2)



Solution

Let’s form an equivalent system whereby the second equation has no x term. First, we can

multiply equation (2) by Ϫ2.

a



2x ϩ 5y ϭ 4

b

Ϫ10x ϩ 14y ϭ 58



(3)

(4)



Now we can replace equation (4) with an equation that we form by multiplying equation (3)

by 5 and then adding that result to equation (4).

a



2x ϩ 5y ϭ 4

b

39y ϭ 78



(5)

(6)



From equation (6) we can find the value of y.

39y ϭ 78

yϭ2

Now we can substitute 2 for y in equation (5).

2x ϩ 5y ϭ 4

2x ϩ 5(2) ϭ 4

2x ϭ Ϫ6

x ϭ Ϫ3

The solution set is {(–3, 2)}.



Classroom Example

5x Ϫ 6y ϭ 3

Solve the system a

b

8x ϩ 4y ϭ 10

using the elimination-by-addition

method.



EXAMPLE 4



Solve the system a



3x Ϫ 2y ϭ 5

b.

2x ϩ 7y ϭ 9



(1)

(2)



Solution

We can start by multiplying equation (2) by Ϫ3.

a



3x Ϫ 2y ϭ

5

b

Ϫ6x Ϫ 21y ϭ Ϫ27



(3)

(4)



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5: Systems of Two Linear Equations

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