80 adult tickets were sold that day.
This is a two equation problem.
First, let:
a = adult tickets
c = child tickets
[tex]6c + 9.6a = 1350[/tex]
[tex]c + a = 177[/tex]
Now, solve for either variable (c or a). We're going to isolate "c" in the second equation and plug it into the first.
[tex]c + a = 177[/tex]
[tex]c = 177 - a[/tex]
Plug it in:
[tex]6(177 - a) + 9.6a = 1350[/tex]
[tex]1062 - 6a + 9.6a = 1350[/tex]
[tex]1062 + 3.6a = 1350[/tex]
[tex]3.6a = 288[/tex]
[tex]a = 80[/tex]
Now, plug in 80 for "a" in either of the equations to find "c."
[tex]c + (80) = 177[/tex]
[tex]c = 97[/tex]
Rather than solving algebraically, as shown above, you can also graph the two equations, and wherever they intersect will be the solution.
