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(39617) (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
and
and
and
Add the opposite of the 900 equation to the 1100 equation.
,
So, what is x?
and then what is z?
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