SOLUTION: solve application problem- system of linear equations in 3 variables x=children $3 y=students $4 z= adult $5 1,000 tickets were sold for a play, which generated $3,800 in

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Question 770163: solve application problem- system of linear equations in 3 variables
x=children $3
y=students $4
z= adult $5
1,000 tickets were sold for a play, which generated $3,800 in revenue. The ticket prices were $3 for children, $4 for students, & $5 for adults. There were 100 fewer student tickets sold than adult tickets. Find the number of each type of ticket sold.
x+y+z=1,000
3x+4y+5z=3,800

y=z-100
y-z=-100
Answer by josgarithmetic(39618) About Me  (Show Source):
You can put this solution on YOUR website!
You have all the right equations. At least try a substitution. Eliminate z through substitution into the first two equations, and then solve for x and y.

Tickets: x+y+z=1,000
Money: 3x+4y+5z=3,800
Adults versus students: z=y+100

x%2By%2By%2B100=1000 and 3x%2B4y%2B5%28y%2B100%29=3800
x%2B2y=900 and 3x%2B9y=3300
x%2B2y=900 and x%2B3y=1100

Add the opposite of the 900 equation to the 1100 equation.
x%2B3y-x-2y=1100-900
highlight%28y=200%29,
So, what is x?
and then what is z?