Question

The local zoo has two water tanks for the elephant exhibit that are leaking. One water tank contains 12 gal of water and is leaking at a constant rate of 3 gal/h. The second water tank contains 8 gal of water and is leaking at a constant rate of 5 gal/h. When will the two tanks have the same amount of water? Explain. Let x = the number of hours the tanks are filling and let y = the number of gallons in the tank. A. They will never have the same amount of water because the solution to the system is (–2,18). It is not possible to have –2 gallons in the tanks. B. They will never have the same amount of water because the solution to the system is (–2,18). It is not possible to have time be –2 hours. C. In 2 hours, because the solution to the system is (2,18). D. In –2 hours, because the solution to the system is (–2,18).

Answer 1

B. They will never have the same amount of water because the solution to the system is (-2,18). It is not possible to have time be -2 hours.

Explanation:

Let us assume that

x = the number of hours the tanks are filling,

y = the number of gallons in the tank.

Tank 1 contains 12 gal of water and is leaking at a constant rate of 3 gal/h.

As the water is decreasing with time, so rate of change will be negative, so the equation will be

`y=12-3x` --------------1

Tank 2 contains 8 gal of water and is leaking at a constant rate of 5 gal/h.

As the water is decreasing with time, so rate of change will be negative, so the equation will be

`y=8-5x` --------------2

As we have to calculate the time where the amount of water left in the tanks will be same, so

`\Rightarrow 12-3x=8-5x`

`\Rightarrow 5x-3x=8-12`

`\Rightarrow 2x=-4`

`\Rightarrow x=-2`

So, at x= -2, y will be

`y=12-3(-2)=12+6=18`

Here, x which is time, is negative i.e solution is not possible.

Also the value of y comes out to be 18, which is impossible. Because there is not even 18 gal of water, so how can they both have 18 gal water.