An approach:

Let X = cost of expensive ticket, Y = cost of cheap ticket, n = number of expensive tickets that can be bought for $1800.

Then:

X = Y + 30 .... (1)

nX = 1800 .... (2)

(n + 10)Y = 1800 => nY + 10Y = 1800 .... (3)

Sub (1) into (2): n(Y + 30) = 1800

Expand: nY + 30n = 1800 .... (4)

(4) - (3) => Y = 3n. Sub this into (3):

(the negative solution is discarded since n > 0).

Sub n = 20 into (2): X = 90.

Sub X = 90 into (1): Y = 60.

Also answered here: http://www.mathhelpforum.com/math-he...-problems.html.

Thread closed.