The small zoo at Nichols Park charges $8 for adult admission, $5 for child admission (18 years old and under), and $3 for senior admission (60 years old and over). For one day, the zoo collected $539 and admitted 96 visitors. If the adult admission is 9 more than the senior admission, how many persons of each category were admitted on this particular day?

To make things a bit simpler, we can represent all three kinds of tickets in terms of one variable, [tex]x[/tex]. Let the number of senior visitors be [tex]x[/tex]. Then, since 9 more adults were admitted as compared to the seniors, we can let the number of adult visitors be [tex]x+9[/tex]. Therefore, since there were 96 visitors in total, we know that the number of child visitors is 96 subtracted by the total number of adult and senior visitors. The total number of adult and senior visitors is [tex]x+x+9=2x+9[/tex]. Thus, the number of child visitors is [tex]96-(2x+9)=96-2x-9=-2x+87[/tex].

Now, all we have to do is write an equation to solve for [tex]x[/tex]. Since senior admission costs $3, we can represent the money earned from it as [tex]3*x=3x[/tex]. Since adult admission costs $8, we can represent the money earned from it as [tex]8*(x+9)=8x+72[/tex]. Finally, since child admission costs $5, we can represent the money earned from it as [tex]5*(-2x+87)=-10x+435[/tex]. Since we know that the zoo collected $539 in total, we can write the following equation to solve for x:

[tex]3x + 8x+72-10x+435=539[/tex]

[tex]x+507=539[/tex] (Simplify LHS)

[tex]x=32[/tex] (Subtract 507 from both sides)

Therefore, 32 seniors were admitted. Since [tex]x=32, x+9=32+9=41[/tex], so 41 adults were admitted, and finally, since [tex]x=32,-2x+87=-2*32+87=-64+87=23[/tex], so 23 children were admitted. Hope this helps!