SOLUTION: hello I cannot seem to figure out the answer to this word problem I don not know what I missing but can someone help me I will greatly appreciate it thank you A basketball t

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Question 884505: hello I cannot seem to figure out the answer to this word problem I don not know what I missing but can someone help me I will greatly appreciate it thank you


A basketball team sells tickets that cost $10, $20, and VIP seats for $30. The team has sold 482 tickets overall. It has sold 132 more $20 tickets than $10 tickets. The total sales are $8850, how many tickets of each kind have been sold?


Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +a+ = number of $10 tickets
Let +b+ = number of $20 tickets
Let +c+ = number of $30 tickets
------------------------------
(1) +a+%2B+b+%2B+c+=+482+
(2) +b+=+a+%2B+132+
(3) +10a+%2B+20b+%2B+30c+=+8850+
-------------------------------
(3) +a+%2B+2b+%2B+3c+=+885+
Multiply both sides of (1) by +3+
and subtract (3) from (1)
(1) +3a+%2B+3b+%2B+3c+=+1446+
(3) +-a+-+2b+-+3c+=+-885+
+2a+%2B+b+=+561+
------------------
Substitute (2) into this result
+2a+%2B+a+%2B+132+=+561+
+3a+=+561+-+132+
+3a+=+429+
+a+=+143+
and
+b+=+a+%2B+132+
+b+=+275+
and
(1) +143+%2B+275+%2B+c+=+482+
(1) +c+=+482+-+418+
(1) +c+=+64+
143 = number of $10 tickets
275 = number of $20 tickets
64 = number of $30 tickets
---------------------------
check:
(3) +10%2A143+%2B+20%2A275+%2B+30%2A64+=+8850+
(3) +1430+%2B+5500+%2B+1920+=+8850+
(3) +8850+=+8850+
OK