Question

William bought some tickets to see his favorite singer. He bought some adults' tickets and some children's tickets, for a total of 30 tickets. Adult's tickets cost $30 per ticket, and children's tickets cost $20 per ticket. If he spent a total of $280, then how many adults and children's tickets did he buy?

Answer 1

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.