Two groups of students order burritos and tacos at a local restaurant. One order of 3 burritos and 4 tacos costs $11.33. the other order of 9 burritos and 5 tacos costs $23.56 ... a. Write a system of equations that describes the situation. b. Solve by elimination to find the cost of a burrito and the cost of a taco.

3 burritos and 4 tacos cost 11.33 3b + 4t = 11.33

9 burritos and 5 tacos cost 23.56 9b + 5t = 23.56

To solve: multiply top equation by 3.

3 (3b + 4t = 11.33) becomes 9b + 12t = 33.99

9b + 5t = 23.56 stays as is 9b + 5t = 23.56

Now subtract the second equation from the first and you will get

7t = 10.43 which you can solve by dividing both sides by 7 to get t = 1.49 so a taco costs $1.49.

Now replace a t in either equation with 1.49 so 3b + 4t = 11.33 becomes 3b + 4 (1.49) = 11.33 or 3b + 5.96 = 11.33. Now subtract 5.96 from each side leaving 3b = 5.37 and finally divide both sides by 3 to get b = 1.79 so a burrito costs $1.79

Problem Two:

3 brass and 10 steel costs $48 3b + 10s = 48

7 brass and 4 steel costs $54 7b + 4s = 54

To solve: multiply top equation by 2 and the bottom equation by 5 so the number of s's will be equal

2 (3b + 10s = 48) becomes 6b + 20s = 96

5 (7b + 4s = 54) becomes 35 b + 20s = 270

Now subtract the top equation from the bottom equation and you will get 29b = 174. Divide both sides by 29 to get b = 6.

Now replace the b of either original equations with the 6 (say the top equation) to get 3 (6) + 10S = 48 or written another way,

10s + 18 = 48. Now subtract 18 from both sides to get 10s = 30 and then divide both sides by 10 to get s = 3.

b=6 and s=3 so each brass part is $6 and each steel part is $3.