1. WARM-UP Date: 3/09/09 1. 4. 3. In a farmyard there are ducks and cows. Ducks have 2 legs and cows have 4 legs. There are 66 legs and a total of 24 animals. How many ducks and cows are there? Solve the system. 2. Solve by graphing. 2x – 3y = -9 -6x + 9y = 30 { 8x – 4y = 16 y = x – 1 { Solve the system. -5x + 6y = 15 4x – 7y = -23 {

2. 3 ( ) 2x – 3y = -9 -6x + 9y = 30 6x -6x + 9y = 30 + 0= 3 Is this true? NO! No Solutions 1. Solve the system. Add systems - 9y= -27

3. 2. Find the solution to the systems by graphing. { 8x – 4y = 16 -8x -4y = 16 – 8x -4 y = -4 + 2x y = x – 1 (3, 2) 8x – 4y = 16 y = x – 1

4. { -2( ) 3. In a farmyard there are chickens and cows. Ducks have 2 legs and cows have 4 legs. There are 66 legs and a total of 24 animals. How many ducks and cows are there? d + c = 24 d: Number of ducks c: Number of cows 2d + 4c = 66 -2d– 2c= -48 2d + 4c = 66 + 2c= 18 2 2 c = 9 9 cows!

5. d + c = 24 d + 9 = 24 c = 9 - 9 d= 15 15 ducks 3. In a farmyard there are chickens and cows. Ducks have 2 legs and cows have 4 legs. There are 66 legs and a total of 24 animals. How many ducks and cows are there?

6. 4 ( ) -5x + 6y = 15 4x – 7y = -23 5 ( ) + 24y-20x 20x + -11y = -55 -11 y = 5 4x – 7y = -23 4x – 7(5) = -23 4x – 35 = -23 + 35 4x = 12 4 4 x = 3 4. Solve the system. Solution (3, 5) Add systems Substitute the x = 60 – 35y= -115

7. Word Problems: Substitution Name: Pd Algebra 3/09/09

8. Ex1: A volleyball club has 41 members. The number of girls is 3 more than the boys. Write a systems of equations representing, b, boys and, g, girls. b + g = 41 { b: Number of boys g: Number of girls g = b + 3

9. Ex2: In 1999, the US produced 20 million more motor vehicles than Japan. Together the two countries produced 22 million vehicles. Write equations to find the number of cars produced in US, u, and Japan, j. u + j = 22 { u: Number of vehicles made in US j: Number of vehicles made in Japan u = j + 20

10. Ex3: Hector and Martha recently collected 32 new stamps. Martha has 2 more than 3 times as many stamps as Hector. Write a system for, h, Hector ’ s stamps, and m, Martha ’ s stamps. m + h = 32 { h: Number of Hector ’ s stamps m: Number of Martha ’ s stamps m = 3h + 2

11. Ex4: John lives 3.5 miles from school. One day he decided to run part of the way home and walk the rest. If the distance John walked is six times as far as he ran, write a system a equations for this situation if, r, is distance ran, and w, is distance walked. w = 6r { r: # of miles ran w: # of miles walked r + w = 3.5

12. Ex5: Maria collects n nickels and d dimes. The number of nickels is 8 less than the number of dimes. If total value of the collection is $1.20, write a system of equations, 0.05n + 0.10d = 1.20 { n: Number of nickels d: Number of dimes n = d – 8

13. Ex6: The perimeter of a given rectangle is 46 inches. If the length is 2 less than five times the width, write a system of equations that represents l, length and w, width. 2w + 2l = 46 { l: Length of rectangle w: Width of rectangle l = 5w – 2

14. Ex7: Ernesto spent a total of $64 for a pair of jeans and a shirt. The jeans cost $6 more than the shirt. What was the cost of the jeans and shirt? j = s + 6 { j: Cost of the jeans s: Cost of the shirt j + s = 64 s + 6 + s = 64 2s+ 6 = 64 - 6 2s = 58 2 2 s = 29 Shirts is $29.

15. j = s + 6 j = 29 + 6 j = 35 s = 29 Jeans are $35. Ex7: Ernesto spent a total of $64 for a pair of jeans and a shirt. The jeans cost $6 more than the shirt. What was the cost of the jeans and shirt?

16. 0.25n + 0.2n Ex8: Jared has $1.55 in dimes and nickels. If the number of of dimes is two less than twice the number nickels. How many dimes does he have? d = 2n – 2 { n:nickels d:dimes 0.05n + 0.1d = 1.55 0.05n + 0.1(2n – 2) = 1.55 0.05n– 0.2 – 0.2 = 1.55 + 0.2 0.25n = 1.75 0.25 n = 7 = 1.55 7 nickels

17. d = 2n – 2 d = 2(7) – 2 d = 12 n = 7 12 dimes Ex8: Jared has $1.55 in dimes and nickels. If the number of of dimes is two less than twice the number nickels. How many dimes does he have?

18. 8w + 6w Ex9: The length of a rectangle is 4 meters less than three times its width. If the perimeter of the rectangle is 48 meters, then what is the value of the length (in meters)? 2W + 2L = 48 { L: length of a rectangle W:width of a rectangle L = 3w – 4 2w + 2(3w – 4) = 48 2w – 8 – 8 = 48 + 8 8w = 56 8 8 w = 7 = 48 Width 7 meters.

19. L = 3(7) – 4 L = 21 – 4 L = 17 w = 7 Length is 17 m. Ex9: The length of a rectangle is 4 meters less than three times its width. If the perimeter of the rectangle is 48 meters, then what is the value of the length (in meters)?