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: [tex]x+y \leq 200[/tex]
One regular ticket cost $12, so [tex]x[/tex] numbers of regular tickets cost [tex]12x[/tex]
One student ticket cost $8, so [tex]y[/tex] numbers of student tickets cost [tex]8y[/tex]
We want to at least get $1000 from the ticket selling, so the inequality is given by: [tex]12x+8y \geq 1000[/tex]
The constraint is given by simplifying both inequalities to the lowest term
First, we have [tex]x+y \leq 200[/tex] ⇒ This is already the lowest term
Second, we have [tex]12x+8y \geq 1000[/tex]
Simplifying this by dividing each term by 4, we have [tex]3x+2y \geq 250[/tex]
The two constraining inequalities are:
[tex]x+y \leq 200[/tex]
[tex]3x+2y \geq 250[/tex]
