Asked by kerry

A basketball team sells tickets that cost​ $10, $20,​ or, for VIP​ seats,​ $30. The team has sold 3401 tickets overall. It has sold 133 more​ $20 tickets than​ $10 tickets. The total sales are​$65960. How many tickets of each kind have been​ sold?
Answered by R_scott
n = $10 , w = $20 , t = $30

n + w + t = 3401 ... 30 n + 30 w + 30 t = 102030

10 n + 20 w + 30 t = 65960

n + 133 = w

subtracting equations (to eliminate t) ... 20 n + 10 w = 36070

substituting ... 20 n + 10 (n + 133) = 36070

solve for n , then substitute back to find w and t
Answered by henry2,
Sold X $10 tickets
Sold x+133 $20 tickets
Sold Y $30 tickets

x + x+133 + y = 3401
Eq1: 2x+y = 3268

10x + 20(x+133) + 30y = 65960
30x+30y = 63300
Eq2: x+y = 2110
Answered by henry2,
Multiply Eq2 by 2 and subtract from Eq1:
2x+y = 3268
2x+2y = 4220
Diff: y = 952 VIP tickets
In Eq1, replace y with 952 and solve for x
2x+952 = 3268
X = 1158 tickets
x+133 = 1158+133 = 1291 tickets
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