SOLUTION: A ballet performance brought in $61,800 on the sale of 3000 tickets. If the tickets sold for $14 and $25, how many of each type of tickets were sold?

Question 132950This question is from textbook Applied College Algebra
: A ballet performance brought in $61,800 on the sale of 3000 tickets. If the tickets sold for $14 and $25, how many of each type of tickets were sold? This question is from textbook Applied College Algebra

Found 2 solutions by checkley71, vleith:
Answer by checkley71(8403) (Show Source): You can put this solution on YOUR website!
14X+25(3000-X)=61,800
14X+75,000-25X=61,800
-11X=61,800-75,000
-11X=-13,200
X=-13,200/-11
X=1,200 TICKETS FOR $14 WERE SOLD.
3,000-1,200=1,800 TICKETS FOR $25 WERE SOLD.
PROOF:
14*1,200+25*1,800=61,800
16,800+45,000=61,800
61,800=61,800

Answer by vleith(2983) (Show Source): You can put this solution on YOUR website!
Given:
Total sales 61,800
Total tickets 3000
two type of tickets, $14 and $25
Total sales = (total of $14) + (total of $25) (Equation 1)
Let x be the number of $14 tickets sold. Then the number of $25 tickets is 3000-x
Sales for $14 tix = $14*x
sales for $25 tix = $25*(3000-x)
Substituting in Equation 1 above:

Yields

Collecting like terms

So x = 1,200 --> we sold 1,200 $14 seats
3000-1200 = 1800. So we sold 1800 $25 seats
Check your answer:
Does 61,800 = 14(1200) + 25(1800)??
yes it does!

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