This is a common version of a system of equations problem. These types of problems show up in algebra and are even asked on the SAT and ACT.
First we'll set the variable x for the number of adult tickets and y for the number of student tickets.
Since adult tickets are $8 and student tickets are $5, we can set the following equation to equal the total amount of money made:
8x + 5y = 1196
Meanwhile, the total number of tickets (190) equals the sum of x and y.
x + y = 190
We always want to get two equations with two variables for these system of equation problems. We'll solve this one using the substitution method, which means we'll take the second equation and isolate one of the variables, in this case x (we could isolate y if we wanted).
x + y = 190, so subtracting y from both sides we get
x = 190 - y
We now take this equation and put it into the first equation for x as follows
8(190 - y) + 5y = 1196
We distribute what's in the parenthesis to get
1520 - 8y + 5y =1196
Now we combine the like terms to get
1520 - 3y = 1196
Subtract 1520 from both sides to get
-3y = -324
Divide both sides by -3 to get
y = 108
That means they sold 108 student tickets. But we're not done yet. We have to figure out how many adult tickets were sold. To do that we substitute 108 for y in the second equation above
x + 108 = 190
We then subtract 108 from each side to get
x = 82
That means we sold 82 adult tickets.
You can always go back and check your answer by multiplying the number of adult tickets by $8 and adding the number of student tickets multiplied by $5 to get the $1196 made in total.
Hope this helps.