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!