To find the day(s) when the attendance was the same at both plays, we need to find the values of x (number of days) for which the attendance y is equal in both equations.
First, let's set the two equations equal to each other:
16x + 150 = -x^2 + 60x - 10
Now we have a quadratic equation. We can rearrange it to get it into the form of ax^2 + bx + c = 0:
x^2 + 44x - 160 = 0
To solve this quadratic equation, we can factor it or use the quadratic formula. In this case, let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values, we have:
x = (-44 ± √(44^2 - 4(1)(-160))) / (2(1))
Simplifying further:
x = (-44 ± √(1936 + 640)) / 2
x = (-44 ± √2576) / 2
x = (-44 ± 50.76) / 2
Now we have two possible values for x:
x = (-44 + 50.76) / 2 = 6.38
x = (-44 - 50.76) / 2 = -47.38
Since the number of days cannot be negative, we can discard the second solution. Therefore, the attendance was the same on day 6.38, which is not a feasible number of days. This means that the attendance was never the same at both plays.
Therefore, the correct answer is:
The attendance was never the same at both plays.