Algebra 1 word problem

Question

Cody and Eugene are selling pies for a school fundraise. Customers can buy blueberry pies and blackberry pies. Cody sold 13 blueberry pies and 8 blackberry pies for a total of $222. Eugene sold 2 blueberry pies and 13 blackberry pies for a toal of $246. Find the cost each of one blueberry pie and one blackberry pie. 

Answer 1

Let B =price for blueberry and K = price for blackberry (had two B's, needed a different letter).

Cody sold 13 B + 8 K = $222

Eugene sold 2 B + 13 K = $246

Two methods to solve the system: substitution and elimination. Neither is straightforward for this problem, but I suggest elimination.

(2)(13 B + 8 K = 222) -> 26 B + 16 K = 444

(-13)(2 B + 13 K = 246) -> -26 B - 169 K = -3198 and add.

-153 K = -2754 so K = -2754/-153 = $18 blackberry

Now can substitute K = 18 (I used 2nd eqn) 2 B + 13(18) = 246 -> 2B + 234 = 246 -> B = 12/2 = 6

K = $18 blackberry, B = $6 blueberry Now can check by substituting into the two original eqns.

Answer 2

13U + 8A = 222

2U +13A = 246

multiply the 1st equation by 2

and multiply the 2nd equation by 13

26U + 169A =13(246) = 3198

26U +16A = 444

subtract to eliminate one variable

153A = 3198-444= 2754

A = 2754/153 = $18 for one BlAckberry Pie

U = (246-13(18))/2 = (246-234)/2 = 12/2 = $6 for one BlUbertry Pie

13(6) + 8(18)= 78+144 = 222

2(6) +13(18) = 12 +234 = 246

$18 for BlAckberry

$6 for BlUberry