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.

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