Water is added to two containers for 16 minutes. The equations below model the ounces of water, y, in each container after x minutes. At the time after the start when the containers hold the same amount of water, how much water do they hold?
Container A: y equals 16 x plus 104
Container B: y equals negative 2 x squared plus 40x plus 160
To find the point at which both containers hold the same amount of water, we need to set the two equations equal to each other:
16x + 104 = -2x^2 + 40x + 160
Rearranging the equation gives us:
2x^2 + 24x + 56 = 0
Now we need to solve this quadratic equation. We can either factor it or use the quadratic formula. Factoring gives us:
2(x^2 + 12x + 28) = 0
2(x + 2)(x + 14) = 0
Setting each factor to 0 to solve for x gives us two possible values:
x + 2 = 0 x + 14 = 0
x = -2 x = -14
Since time cannot be negative, we discard the -14 as the extraneous solution. Therefore, x = -2, but since we're looking for the time after the start when the containers have the same amount of water, we consider x = 2.
Plugging x = 2 into either of the original equations gives us:
y = 16(2) + 104
y = 32 + 104
y = 136
Therefore, at 2 minutes after the start, both containers hold 136 ounces of water.