1. Elimination Using Addition and Subtraction CFU 3102.3.28 Solve systems of linear equations graphically, algebraically, and with technology.

2. The cost for renting a car from Shady Grady Rent-a- Car is $18 per day plus 30¢ per mile driven. The cost of renting a similar car from EZ-Rider Rental is $20 per day plus 25¢ per mile driven. Cathy needs to rent a car for one day. Should she rent from Shady Grady or EZ-Rider? Use c for cost and m for miles driven. Write an algebraic expression fore each car rental option. Solve the system to determine the best option. Turn your homework(worksheet on solving equations by substitution) into the box

3. The cost for renting a car from Shady Grady Rent-a-Car is $18 per day plus 30¢ per mile driven. The cost of renting a similar car from EZ- Rider Rental is $20 per day plus 25¢ per mile driven. Cathy needs to rent a car for one day. Should she rent from Shady Grady or EZ-Rider? SG C = 18 + 0.3m EZ C = 20 + 0.25m It depends, look at the graph

4. There are also fixed expenses for operating a car, such as loan repayment and insurance. The weekly fixed expenses are $40 for Car A, and $55 for Car B. This information can be used to draw a graph that shows the total weekly operating costs for the two cars. Each line on the graph at the right has the same slope as the corresponding line on the graph above. However, the intercept of each line is the fixed cost per week. Study the graph. The operating cost for Car A is less when a person drives less than 600 miles per week. When is the operating cost for Car B less than for Car A? Write an algebraic expression fore each car

5. Elimination Method - Addition 3x – 5y = -16 2x + 5y = 31 5x = 15 (-5y and 5y cancel) x = 3 (divide by 5) 3(3) – 5y = -16 (substitute) 9 – 5y = -16 (move the 9) -5y = -16 – 9 -5y = -25 (divide by -5) y = -5 (3, -5) solution Process 1.Add the two problems together to “eliminate” one of the variables 2.Use substitution method to finish

6. Elimination Method - Subtraction 5s + 2t = 6 9s + 2t = 22 (subtract from first) -4s + 0 = -16 -4s = - 16 divide by – 4 s = -4 substitute into equation 5(-4) + 2t = 6 -20 + 2t = 6 move 20 2t = 6 + 20 2t = 26 divide by 2 t = 13 (-4, 13) answer Process 1.Multiple equation by -1 change sign to be opposite. 2.Add the two problems together to “eliminate” one of the variables 3.Use substitution method to finish

7. Which Variable will you eliminate? -3x + 4y = 12 3x – 6y = 18 8x + y = 16 -6x + y = -26 s – t = 4 s + t = 2 3a + b = 5 2a + b = 10 2/3x – ½ y = 14 5/6 x – ½ y = 18 3x Add Y (subtract)T (add) b (subtract) Y (subtract)

8. Solving Systems by Adding x – y = 1 x + y = 3 2x = 4 x = 2 2 – y = 1 -y = 1 – 2 -y = -1 -Y = 1 (2, -1) -x + y = 1 x + y = 11 2y = 12 y = 6 x + 6 = 11 x = 11 – 6 x = 5 (5, 6)