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.
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
1 answer