Final answer:
William did not buy any valid combination of adults' and children's tickets.
Step-by-step explanation:
To solve this problem, we can set up a system of equations to represent the given information:
Let x be the number of adults' tickets that William bought.
Let y be the number of children's tickets that William bought.
We can then write the following equations:
x + y = 30 (equation 1)
30x + 20y = 280 (equation 2)
We can solve this system of equations using the method of substitution or elimination. Let's use the method of substitution:
From equation 1, solve for x:
x = 30 - y
Substitute the value of x in equation 2:
30(30 - y) + 20y = 280
Expand and simplify:
900 - 30y + 20y = 280
Combine like terms:
900 - 10y = 280
Subtract 900 from both sides:
-10y = -620
Divide both sides by -10:
y = 62
Substitute the value of y into equation 1:
x + 62 = 30
Subtract 62 from both sides:
x = -32
Since the number of tickets cannot be negative, we discard the solution x = -32. Therefore, William bought 62 children's tickets and 30 - 62 = -32 adults' tickets, which is not a valid solution.
Therefore, there is no valid solution to this problem.