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Let's first analyze the equation in the problem: 

\[
3200 + 25x = 2800 + 45x.
\]

This equation models a situation where two pools, Pool A and Pool B, have different starting amounts of water and different rates at which water is added.

### Step-by-step breakdown:
- **Pool A** starts with 3200 liters and is filled at a rate of 25 liters per minute.
- **Pool B** starts with 2800 liters and is filled at a rate of 45 liters per minute.

The equation represents when the amount of water in Pool A equals the amount of water in Pool B after *x* minutes.

### Solving the equation for *x*:
\[
3200 + 25x = 2800 + 45x.
\]
First, subtract 2800 from both sides:
\[
400 + 25x = 45x.
\]
Next, subtract 25x from both sides:
\[
400 = 20x.
\]
Finally, divide by 20:
\[
x = \frac{400}{20} = 20.
\]

Thus, **x = 20 minutes**, meaning after 20 minutes, the amount of water in both pools will be equal.

### Answer for part (b):
For the equation, **x = 20** minutes.

Would you like more details on this problem, or do you have any questions?

Here are 5 related questions to expand the discussion:
1. What would happen if Pool A filled at 30 liters per minute instead of 25 liters per minute?
2. How would the equation change if Pool B started with 3000 liters instead of 2800 liters?
3. How many liters of water will each pool contain after 20 minutes?
4. How can you represent this situation with a graph?
5. What would the equation look like if water was removed from one of the pools instead of added?

**Tip**: When solving linear equations, it's often helpful to isolate the variable (in this case, *x*) by performing inverse operations step-by-step.

3200 + 25x = 2800 + 45x. Solve for x based on water filling rates for two pools.

Solve the linear equation 3200 + 25x = 2800 + 45x. This problem involves water filling rates for two pools, using algebraic concepts to find when both pools have the same amount of water.

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