So for this we'll be using the system of equations. The first equation is gonna be representing the cost of the tickets and the second equation is gonna represent the number of adult tickets sold.
Let x equal # of adult tickets and y equal # of student tickets
The first equation is [tex]8x+4y=880[/tex]
The second equation is [tex]x=y+20[/tex]
For this, I'll be using the substitution method. And to do that, we can substitute x in the first equation for (y+20) since that is the equivalent according to the second equation.
And so your first equation should look like this: [tex]8(y+20)+4y=880[/tex] . And from there we can solve.
Multiply 8 and (y+20), and your equation should look like [tex]8y+160+4y=880[/tex]
Combine like terms to get [tex]12y+160=880[/tex]
Subtract 160 on each side to get [tex]12y=720[/tex]
And then divide 12 on each side and your answer should be [tex]y=60[/tex]
Now that we know that y=60, we can substitute y for 60 in either equation to solve for x. For this, i'll be using the second equation.
[tex]x=60+20 \\ x=80[/tex]
The drama class sold 80 adult tickets.
