SOLUTION: A family purchased 6 pizzas and 4 pitchers of soda for $114. Another family purchased 9 pizzas and 7 pitchers of soda for $175.50. How much for each pizza and soda?

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Question 1003212: A family purchased 6 pizzas and 4 pitchers of soda for $114. Another family purchased 9 pizzas and 7 pitchers of soda for $175.50. How much for each pizza and soda?
Found 2 solutions by fractalier, josmiceli:
Answer by fractalier(6550) About Me  (Show Source):
You can put this solution on YOUR website!
Let the prices of pizza and soda be x and y. Then we have
6x + 4y = 114
9x + 7y = 175.5
Multiply the top equation by three and the bottom one by two and subtract them...we get
18x + 12y = 342
-(18x + 14y = 351)
and
-2y = -9
y = $4.50
Substituting in to the first equation we get
6x + 4(4.50) = 114
6x + 18 = 114
6x = 96
x = $16.00

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
+6p+%2B+4s+=+114+
+9p+%2B+7s+=+175.5+
-------------------
(1) +3p+%2B+2s+=+57+
(2) +18p+%2B+14s+=+351+
----------------------
Multiply both sides of (1) by +6+ and
subtract (1) from (2)
(2) +18p+%2B+14s+=+351+
(1) +-18p+-+12s+-342+
-----------------------
+2s+=+9+
+s+=+4.5+
and
(1) +3p+%2B+2%2A4.5+=+57+
(1) +3p+%2B+9+=+57+
(1) +3p+=+48+
(1) +p+=+16+
Pizzas are $16 ea
Sodas are $4.50 ea