algebra word problem

Question

For a school play, student tickets cost 40 cents and adults tickets were 75 cents. The total receipts for 90 tickets were $46.50. How many of each kind were sold?

Answer 1

let x =  number of student tickets, and y = number of adult tickets

 

then x + y = 90 tickets

 

then x tickets times $0.40/student ticket  + y tickets times $0.75/adult ticket = $46.50; this equation becomes $0.40x + $0.75y = $46.50

 

we solve the first equation for x, which gives us x = 90t - y

 

we substitute that value into the second equation, which gives us

$0.40(90 - y) + $0.75y = $46.50, which becomes $36.00 - $0.40y + $0.75y = $46.50

 

after combining terms, we have $0.35y = $10.50; so y = $10.50 / $0.35 = 30 tickets

 

thus x = 90 tickets - 30 tickets = 60 tickets

 

Proof: 60t x $0.40/t + 30t x $0.75/t = $46.50; $24.00 + $22.50 = $46.50