Question 162558: a basketball team sells tickets that cost $10 $20 or VIP seats, $30. The team has sold 492 tickets over all. It has sold 177 more $20 tickets than $10 tickets. The total sales are $8580. How Many tickets of each kind have been sold?
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! A basketball team sells tickets that cost $10 $20 or VIP seats, $30.
The no. of three types of tickets = x, y, z
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The team has sold 492 tickets over all.
x + y + z = 492
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It has sold 177 more $20 tickets than $10 tickets.
y = x + 177
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The total sales are $8580.
10x + 20y + 30z = 8580
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How Many tickets of each kind have been sold?
:
Replace (x+177) for y in the 1st and 3rd equations
x + (177+x) + z = 492
2x + z = 492 - 177
2x + z = 315
and
10x + 20(x+177) + 30z = 8580
10x + 20x + 3540 + 30z = 8580
30x + 30z = 8580 - 3540
30x + 30z = 5040
Simplify, divide equation by 30
x + z = 168
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Using these two equations for elimination
2x + z = 315
x + z = 168
---------------subtracting eliminates z
x = 147 ea $10 tickets sold
:
y = 147 + 177
y = 324 ea $20 tickets sold
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Find z using the 1st equation
147 + 324 + z = 492
471 + z = 492
z = 492 - 471
z = 21 ea $30 tickets sold
:
Check solution in the $$ equation
10(147) + 20(324) + 30(21) =
1470 + 6480 + 630 = 8580; confirms our solutions
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