Let x be the number of regular tickets
Let y be the number of student tickets
At the most, we can seat 200 people (both regular tickets and student tickets holder). So the inequality is given by: \(x+y \leq 200\)
One regular ticket cost $12, so @@MATH_TOKEN_1@@ numbers of regular tickets cost @@MATH_TOKEN_2@@
One student ticket cost $8, so \(y\) numbers of student tickets cost \(8y\)
We want to at least get $1000 from the ticket selling, so the inequality is given by: \(12x+8y \geq 1000\)
The constraint is given by simplifying both inequalities to the lowest term
First, we have \(x+y \leq 200\) ⇒ This is already the lowest term
Second, we have \(12x+8y \geq 1000\)
Simplifying this by dividing each term by 4, we have \(3x+2y \geq 250\)
The two constraining inequalities are:
\(x+y \leq 200\)
\(3x+2y \geq 250\)
