Chapter 8Systems of Linear Equations

in Two Variables

Section 8.2

Chapter 8Systems of Linear Equations in Two Variables

Section 8.2

Section 8.2Exercise #17

Chapter 8Systems of Linear Equations in Two Variables

4x – y = – 1

2x + 4y = 13

Solve using the substitution method.

4x – y = – 1

2x + 4y = 13

2x + 4 4x + 1 = 13

2x + 16x + 4 = 13

18x = 9

x =

918

= 12

y = 4x + 1

– y = – 4x – 1

Solve using the substitution method.

y = 4

12

+ 1

4x – y = – 1

– y = – 4x – 1

y = 2 + 1

y = 3

12

, 3

Solve using the substitution method.

Section 8.2Exercise #23

Chapter 8Systems of Linear Equations in Two Variables

Solve using the substitution method.

y = –

32

x + 4

3x + 2 –

32

x + 4

= – 1

3x + – 3x + 8 = – 1

8 = – 1No solution.

3x + 2y = – 1

32

x + y = 4

Section 8.2Exercise #25

Chapter 8Systems of Linear Equations in Two Variables

10x – 30y = – 10 2x – 6y = – 2

Solve using the substitution method.

Solve using the substitution method.

2x – 6y = – 2

x = 3y – 1 2x = 6y – 2

10x – 30y = – 10

Solve using the substitution method.

2x – 6y = – 2

x = 3y – 1 2x = 6y – 2

– 10 = – 10

Infinitely many solutions.

10 3y – 1 – 30y = – 10

30y – 10 – 30y = – 10

Solve using the substitution method.

y =

– 2 – 6

x – 2

– 6

– 6y = – 2x – 2

y =

13

x + 13

2x – 6y = – 2

x , y y =

13

x + 13

Infinitely many solutions of the form

Section 8.2Exercise #51

Chapter 8Systems of Linear Equations in Two Variables

Two angles are complementary. One angle is 10º more than 3 times the other angle. Find the measure of each angle. x = one angle y = second angle

x + y = 90

4y + 10 = 90

x =

10 + 3y + y = 90

10 + 3y

4y = 80

y = 20

x = 10 + 3 20

Two angles are complementary. One angle is 10º more than 3 times the other angle. Find the measure of each angle. x = one angle

x + y = 90

x = 10 + 60

x = 70

The angles are 70 and 20 .

y = second angle

Section 8.2Exercise #57

Chapter 8Systems of Linear Equations in Two Variables

Two water purification systems are priced differently.Company A charges a $55 installation fee plus $20 permonth. Company B charges $22.50 per month with noinstallation fee. The total cost, y , depends on the numberof months, x , according to the following equations:

Company A: y = 20x + 55

800 700 600 500 400 300 200 100

0 0 x

y

5 10 15 20 25 30 35

Cost

($)

Number of Months

Total Cost to Rent a Water Purification

System

y = 22.5x

y = 20x + 55

Company B: y = 22.50x

a. Use the graph to determine how many months are required for the cost of the rent from Company A to equal the cost to rent from Company B.

22 months

800 700 600 500 400 300 200 100

0 0 x

y

5 10 15 20 25 30 35

Cost

($)

Number of Months

Total Cost to Rent a Water Purification

System

y = 22.5x

y = 20x + 55

b. Use the substitution method to solve the system of equations. Interpret the answer in terms of the number of months and the total cost of renting a water purification system.

y = 20x + 55

y = 22.50x 22.50x = 20x + 55

2.50x = 55

x = 22 y = 22.50 22 y = 495

Solution: 22, 495

After renting a system for 22 months,both companies will charge $495.

c. Which company is more expensive if the water system is rented for more than 22 months? Which is more expensive if the water system is rented for less than 22 months?

Company B is more expensive thanA for more than 22 months. CompanyA is more expensive than B for lessthan 22 months.