SOLUTION: Tickets for a play at the community theater cost $18 for an adult and $10 for a child. If 80 tickets were sold and the total receipts were $1040, how many of each type of ticket w
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Question 145195: Tickets for a play at the community theater cost $18 for an adult and $10 for a child. If 80 tickets were sold and the total receipts were $1040, how many of each type of ticket were sold?
Found 2 solutions by jojo14344, nerdybill:
Answer by jojo14344(1513) (Show Source): You can put this solution on YOUR website!
Okay, let "x"= adult's ticket, and "y"= child's ticket
since 80 tickets were sold, "x+y = 80" right? ---------- eqn 1
We know the cost for each adult's ticket and also for the child that totals $1040. Putting that into equation,
$18(x) + $10(y) = $1040 --------------- eqn 2
In eqn 1 we get, "y=80-x", and substitute this in eqn 2:
18(x)+ 10(80-x) = 1040
18x+800-10x=1040
8x=1040-800
x=30 ---------------- This the number of adult's ticket sold
For y, go back eqn 1,
30+y=80
y=80-30=50 ---------------- This the number of child's ticket sold
In doubt? Go back eqn 2,
18(30)+10(50) = 1040
540+500=1040
$1040=$1040
Thank you,
Jojo
Answer by nerdybill(7384) (Show Source): You can put this solution on YOUR website!
Tickets for a play at the community theater cost $18 for an adult and $10 for a child. If 80 tickets were sold and the total receipts were $1040, how many of each type of ticket were sold?
.
Let x = adult tickets sold
and y = child tickets sold
then
x + y = 80
18x + 10y = 1040
.
solving eq 1 for x:
x + y = 80
x = 80 - y
.
plug the value of x above into eq 2 and solve for y:
18x + 10y = 1040
18(80 - y) + 10y = 1040
1440 - 18y + 10y = 1040
1440 - 8y = 1040
-8y =-400
y = 50
.
plug the value of y above into eq 1 and solve for x:
x + y = 80
x + 50 = 80
x = 30
.
Tickets sold: 30 adult and 50 child
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