Chapter 8 Section 4Solving System of Equations

Applications and Problem Solving

Learning Objective

Use systems of equations to solve application problems

Key Vocabulary: complementary anglessupplementary angles

Solve Systems of EquationsApplication and Problem Solving

1. Complementary angle is two angles whose sum measures 90◦ x + y = 90

2. Supplementary angle is two angles whose sum measures 180◦ x + y = 180

3. Perimeter of a Rectangle P = 2L + 2W

4. Interest = Principle * rate * time i = prt

5. Distance = rate * time d = rt

Example: Equation 1

A + B = 90

Equation 2A = B + 26A – B = 26

Angles A and B are complementary angles. If angle A is 26◦ greater than angle B. Find the measure of each angle.

Step 1:Variables are already on one side and constants are on the other side

Step 2multiply one or both equations by a constant(s)

Step 3Add the equations resulting in one equation

A + B = 90 A – B = 262A = 116

Step 4Solve for the variable

2A = 116A = 116/2A = 58

A + B = 9058 + B = 90B = 90 – 58 B = 32

Step 5Substitute the value found into one of the original equations, and solve that equation for the other variable

A = 58B = 32

Example: Equation 1

2L + 2W = 92

Equation 2L = W + 6L – W = 6

2(L – W = 6)2L – 2W = 12

Determine the dimensions of a rectangle having a perimeter of 92 meters and whose length is 6 meters greater than the width.

Step 1:Variables are already on one side and constants are on the other side

Step 2multiply one or both equations by a constant(s)

Step 3Add the equations resulting in one equation

2L + 2W = 92 2L – 2W = 124L = 104

Step 4Solve for the variable

4L = 104L = 104/4L = 26

2(26) + 2W = 9252 + 2W = 922W = 92 – 52 2W = 40W = 20

Step 5Substitute the value found into one of the original equations, and solve that equation for the other variable

L = 26W = 20

Example: Equation 1

C + A = 110

Equation 2C(1) + A(1.5) = 140C + 1.5 A = 140

-1(C + A = 110)-C – A = -110

Raffle tickets cost $1.00 for children and $1.50 for adults. If 110 tickets where sold for a total of $140.00 How many adults tickets and how many children tickets were sold?

Step 1:Variables are already on one side and constants are on the other side

Step 2multiply one or both equations by a constant(s)

Step 3Add the equations resulting in one equation

C + 1.5 A = 140 -C – A = -110.5 A = 30

Step 4Solve for the variable

.5 A = 30A = 60

C + 60 = 110C = 110 – 60 C = 50

Step 5Substitute the value found into one of the original equations, and solve that equation for the other variable

C = 50A = 60

Example: Equation 1

C = 800 + 25n

Equation 2C = 200 + 30n

800 + 25n = 200 + 30n800 – 200 = 30n – 25n 600 = 5n120 = n

John and Rosemary are considering two halls to rent for their wedding reception. One hall charges $800 plus $25 per person for food. The second hall charges $200 plus $30 per person for food. How many people are needed for the total cost of both halls to be the same?

120 people are needed for the cost to be the same.

Example: Equation 1

x = y + 10

Equation 25x + 5y = 600

5(y + 10) + 5y = 6005y + 50 + 5y = 60010 y = 600 – 5010 y = 550y = 550/10y = 55

x = 55 + 10x = 65

Two people travel in opposite directions on an interstate. They leave at the same time with one person traveling 10 mph faster than the other. Determine both speeds if they are 600 miles apart 5 hours later.

Person 1 = 65 mphPerson 2 = 55 mph

Rate Time Distance

Person 1 x 5 5x

Person 2 y 5 5y

Example: Equation 1

x + y = 25x = -y + 25

Equation 22.5x + 3y = 2.8(25)2.5x + 3y = 70

2.5(-y + 25) +3y= 70-2.5 y + 62.5 + 3y = 70.5 y = 70 – 62.5.5 y = 7.5y = 15

Peanuts sell for $2.50 per pound and walnuts sell for $3.00 per pound. How many pound of each should be used to obtain a mixture of 25 pounds that sell for $2.80 per pound?

x + y = 25x + 15 = 25x = 25 – 15 x = 10

Peanuts = x = 10 lbsWalnuts = y = 15 lbs

Nut Price # lbs Value

Peanuts 2.50 x 2.50x

Walnuts 3.00 y 3.00y

Mix 2.80 25 2.80(25)

Example: Equation 1

x + y = 30x = 30 - y

Equation 20.05x + 0.20y = 30(0.10)0.05x + 0.20y = 3

0.05x + 0.20y = 3 0.05(30 - y) + 0.20y = 31.5 – 0.05y + 0.20y = 3.15 y = 3 - 1.5.15y = 1.5y = 10

How many liters of a 5% solution and a 20% solution should be combined to obtain 30 liters of a 10% solution?

x + y = 30x + 10 = 30x = 30 – 10 x = 20

20 L of 5%10 L of 20%

Liters Concentration Content

5% x 0.05 0.05x

20% y 0.20 0.20y

Mix 30 0.10 0.1(30)

Remember

Draw or label diagrams whenever possible to help vision your equation and results.

You should always check your answers.

It may be helpful with application problem if you work together.

Always read then re-read to make sure you understand what is being ask.

HOMEWORK 8.4

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