One hose can fill a pool 1.5 times as fast as a second hose. When both hose are opened, they fill the pool in four hours. How long would it take to fill the pool if only the slower hose is used?

Step 1 ：Let's denote the time it takes for the slower hose to fill the pool as \(x\) hours.

Step 2 ：The faster hose can fill the pool 1.5 times as fast as the slower one, so it would take \(\frac{2}{3}\) of the time of the slower hose to fill the pool, which is \(\frac{2}{3}x\) hours.

Step 3 ：The rate of work done by the slower hose is \(\frac{1}{x}\) (pool per hour) and the rate of work done by the faster hose is \(\frac{1}{\frac{2}{3}x} = \frac{3}{2x}\) (pool per hour).

Step 4 ：When both hoses are opened, they can fill the pool in 4 hours. So, their combined rate is \(\frac{1}{4}\) (pool per hour).

Step 5 ：Therefore, we can set up the equation: \(\frac{1}{x} + \frac{3}{2x} = \frac{1}{4}\).

Step 6 ：To solve this equation, we first find a common denominator, which is \(4x\): \(\frac{4}{4x} + \frac{6}{4x} = \frac{1}{4}\).

Step 7 ：Combine like terms: \(\frac{10}{4x} = \frac{1}{4}\).

Step 8 ：Cross multiply: \(10 = x\).

Step 9 ：So, it would take 10 hours to fill the pool if only the slower hose is used. This is our final answer: \(\boxed{10}\).

Step 10 ：To check our answer, we can substitute \(x = 10\) into the original equation: \(\frac{1}{10} + \frac{3}{2*10} = \frac{1}{4}\).

Step 11 ：Simplify: \(0.1 + 0.15 = 0.25\).

Step 12 ：The left-hand side equals the right-hand side, so our answer is correct.