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?

Question

Container A: y=16x+104
Container b: y=-2x^2+40x+160

Answer 1

To find the amount of water the containers hold when they have the same amount, we need to set their y values equal to each other and solve for x.

16x + 104 = -2x^2 + 40x + 160

Rearrange the equation to set it equal to zero:

2x^2 - 24x - 56 = 0

Next, factor the quadratic equation:

2(x^2 - 12x - 28) = 0
2(x - 14)(x + 2) = 0

Set each factor to zero to solve for x:

x - 14 = 0
x = 14

or

x + 2 = 0
x = -2

Since time cannot be negative, we discard the negative solution and use x = 14.

Now plug x = 14 into either of the original equations to find the amount of water in each container:

For Container A: y=16(14)+104 = 248 ounces
For Container B: y=-2(14)^2+40(14)+160 = 248 ounces

Therefore, when the containers hold the same amount of water, they hold 248 ounces.