algebra systems

Question

Two pools are being filled with water. To start, the first pool contains 707 liters of water and the second pool is empty. Water is being added to the first pool at a rate of 20.25 liters per minute. Water is being added to the second pool at a rate of 45.5

 liters per minute.

After how many minutes will the two pools have the same amount of water?

How much water will be in each pool when they have the same amount?

Answer 1

In this problem, we know that the two pools will wind up having the same amount of water.

1. The first pool already has 707 liters, and more water is being added at a rate of 20.25 liters per minute.
2. So the expression that describes the situation for Pool #1 is 707 + 20.25x, or 20.25x + 707, where the variable "x" is the number of minutes.
3. The second pool is empty, and water is being added at a rate of 45.5 liters per minute.
4. The expression that describes the situation for Pool #2 is 45.5x, again where "x" is the number of minutes.
5. Since we know that both pools will eventually contain the same amount of water, we can set these two expressions equal to each other.
6. Therefore, 45.5x = 20.25x + 707
7. Combining like terms, we subtract 20.25x from each side of the equation. This gives us 25.25x = 707
8. From here we can isolate the variable by dividing each side of the equation by 25.25. This leaves us with x = 28 minutes. So that's the answer to the first part of the problem.
9. To find out how much water is in each pool, we simply substitute our answer of 28 minutes into the original expressions. They should be equal to each other..
10. Completing this step gives us 45.5(28) = 1,274 liters, and 20.25(28) + 707 = 1,274 liters.

In summary, after 28 minutes both pools will contain 1,274 liters of water.

Answer 2

707 + 20.25m = 0 + 45.5m

Can you solve for m and answer?