1. Chapter 8 Section 4 Solving System of Equations Applications and Problem Solving

2. Learning Objective Use systems of equations to solve application problems Key Vocabulary: complementary angles supplementary angles

3. Solve Systems of Equations Application 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 * timei = prt 5.Distance = rate * timed = rt

4. Example: Equation 1 A + B = 90 Equation 2 A = B + 26 A – 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 2 multiply one or both equations by a constant(s) Step 3 Add the equations resulting in one equation A + B = 90 A – B = 26 2A = 116 Step 4 Solve for the variable 2A = 116 A = 116/2 A = 58 A + B = 90 58 + B = 90 B = 90 – 58 B = 32 Step 5 Substitute the value found into one of the original equations, and solve that equation for the other variable A = 58 B = 32

5. Example: Equation 1 2L + 2W = 92 Equation 2 L = W + 6 L – 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 2 multiply one or both equations by a constant(s) Step 3 Add the equations resulting in one equation 2L + 2W = 92 2L – 2W = 12 4L = 104 Step 4 Solve for the variable 4L = 104 L = 104/4 L = 26 2(26) + 2W = 92 52 + 2W = 92 2W = 92 – 52 2W = 40 W = 20 Step 5 Substitute the value found into one of the original equations, and solve that equation for the other variable L = 26 W = 20

6. Example: Equation 1 C + A = 110 Equation 2 C(1) + A(1.5) = 140 C + 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 2 multiply one or both equations by a constant(s) Step 3 Add the equations resulting in one equation C + 1.5 A = 140 -C – A = -110.5 A = 30 Step 4 Solve for the variable.5 A = 30 A = 60 C + 60 = 110 C = 110 – 60 C = 50 Step 5 Substitute the value found into one of the original equations, and solve that equation for the other variable C = 50 A = 60

7. Example: Equation 1 C = 800 + 25n Equation 2 C = 200 + 30n 800 + 25n = 200 + 30n 800 – 200 = 30n – 25n 600 = 5n 120 = 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.

8. Example: Equation 1 x = y + 10 Equation 2 5x + 5y = 600 5(y + 10) + 5y = 600 5y + 50 + 5y = 600 10 y = 600 – 50 10 y = 550 y = 550/10 y = 55 x = 55 + 10 x = 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 mph Person 2 = 55 mph RateTimeDistance Person 1x55x Person 2y55y

9. Example: Equation 1 x + y = 25 x = -y + 25 Equation 2 2.5x + 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 = 7.5 y = 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 = 25 x + 15 = 25 x = 25 – 15 x = 10 Peanuts = x = 10 lbs Walnuts = y = 15 lbs NutPrice# lbsValue Peanuts2.50x2.50x Walnuts3.00y3.00y Mix2.80252.80(25)

10. Example: Equation 1 x + y = 30 x = 30 - y Equation 2 0.05x + 0.20y = 30(0.10) 0.05x + 0.20y = 3 0.05(30 - y) + 0.20y = 3 1.5 – 0.05y + 0.20y = 3.15 y = 3 - 1.5.15y = 1.5 y = 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 = 30 x + 10 = 30 x = 30 – 10 x = 20 20 L of 5% 10 L of 20% LitersConcentrationContent 5%x0.050.05x 20%y0.200.20y Mix300.100.1(30)

11. 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.

12. HOMEWORK 8.4 Page 520 - 521 #1, 3, 5, 7, 11