For this problem I set up two equations, one for each student.

For Janis I have: 6a + 15s = 48 where a represents price of adult tickets and s represents price of student tickets.

For Tom I have: 8a + 7s = 38 where a and s represent the same thing as before.

With these two equations we can use substitution to solve for a variable.

I chose to solve for a in Tom's equation.

8a + 7s = 38 (original equation)

8a = 38 - 7s (subtract 7s from both sides)

a = 4.75 - 0.875s (divide both sides by 8)

Now we can plug this new equation for a into Janis's original equation.

6a + 15s = 48 (original equation)

6(4.75 - 0.875s) +15s = 48 (substitute equation for a)

28.5 - 5.25s + 15s = 48 (distribute)

28.5 + 9.75s = 48 (combine like terms)

9.75s = 19.5 (subtract 28.5 from both sides)

s = 2 (divide both sides by 9.75)

Now we can substitute our value for s into the equation we found for a.

a = 4.75 - 0.875s (original equation)

a = 4.75 - 0.875(2) (substitute s = 2)

a = 4.75 - 1.75 (multiply 0.875 and 2)

a = 3 (subtract 4.75 - 1.75)

We now get that a = 3 and s = 2.

Since we know that these represent the value of the tickets, we know that adult tickets cost $3 and student tickets cost $2.

Hope this helps!