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 16x + 104 container b y equals negative 2x + 40x + 160
To find the point in time when both containers hold the same amount of water, we need to find the values of X that make Y equal in both equations.
Set the equations equal to each other:
16x + 104 = -2x + 40x + 160
Combine like terms:
16x + 104 = 38x + 160
Subtract 16x from both sides:
104 = 22x + 160
Subtract 160 from both sides:
-56 = 22x
Divide by 22:
x = -56/22
x = -28/11
The containers hold the same amount of water after approximately 28/11 minutes.
To find out how much water they hold at that time, substitute x back into one of the equations. Let's use container A:
y = 16(-28/11) + 104
y = -448/11 + 104
y ≈ 56/11
Therefore, both containers hold approximately 56/11 ounces of water after about 28/11 minutes.