3 variable word problem. help me solve ^_^?

Question

Last Tuesday, Real Cinemas sold a total of 8,500 movie tickets. Proceeds totaled $64,600. Tickets can be bought in one of 3 ways: a matinée admission costs $5, student admission is $6 all day, and regular admissions are $8.50. How many of each type of ticket was sold if twice as many were sold as matinée tickets?

ok this is what i got so far but I dunno if it is right.

x+y+z=8,500
5x+6y+8.50z=64,000
2x-y=0z

THANKS for the help if you help me out :)

"How many of each type of ticket was sold if twice as many student tickets were sold as matinée tickets"

i ment that yeah =p

Answer 1

You left out something in the last sentence; I assume from your third equation that the sentence should read "How many of each type of ticket was sold if twice as many student tickets were sold as matinée tickets?"

If my assumption is correct, then your equations are correct, except that the right side of the second equation should be 64600.

I would solve by substitution, because the last equation has only two variables in it. Solve it for y in terms of x (it's y = 2x, right?)

I would then use that to substitute for y in the first equation: you get x + (2x) + z = 8500, or 3x + z = 8500. Solve this for z in terms of x. (You get z = 8500 - 3x, right?)

Finally, I would substitute using y=2x and z=8500-3x in the second equation to get an equation in x only. Once you get the solution for x, use it in y=2x to get y and in z=8500-3x to get z.