9. A water tank can be filled with water by two pipes in 1 hour and 20 minutes. If the smaller pipe takes 2 hours longer than the larger pipe to fill the tank, find the time taken by each pipe to fill the tank.

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Let's denote the time taken by the larger pipe to fill the tank as $$ L $$ hours and the time taken by the smaller pipe as $$ S $$ hours.

Given that the smaller pipe takes 2 hours longer than the larger pipe to fill the tank, we have the equation:
$$ S = L + 2 $$

The combined rate of the two pipes filling the tank is the sum of their individual rates. The rate of the larger pipe is $$ \frac{1}{L} $$ tanks per hour, and the rate of the smaller pipe is $$ \frac{1}{S} $$ tanks per hour.

The combined rate of the two pipes is $$ \frac{1}{1 \text{ hour and } 20 \text{ minutes}} $$ tanks per hour. We can convert 1 hour and 20 minutes to just hours by dividing 20 minutes by 60 minutes and adding it to 1 hour:
$$ 1 \text{ hour and } 20 \text{ minutes} = 1 + \frac{20}{60} = 1 + \frac{1}{3} = \frac{4}{3} \text{ hours} $$

Therefore, the combined rate of the two pipes is $$ \frac{1}{\frac{4}{3}} = \frac{3}{4} $$ tanks per hour.

We can set up the equation based on the combined rate:
$$ \frac{1}{L} + \frac{1}{S} = \frac{3}{4} $$

Substitute $$ S = L + 2 $$ into the equation:
$$ \frac{1}{L} + \frac{1}{L + 2} = \frac{3}{4} $$

Now, we can solve this equation to find the time taken by each pipe to fill the tank.
Solve the equation by following steps:
- step0: Solve for $$L$$:
  $$\frac{1}{L}+\frac{1}{L+2}=\frac{3}{4}$$
- step1: Find the domain:
  $$\frac{1}{L}+\frac{1}{L+2}=\frac{3}{4},L \in \left(-\infty,-2\right)\cup \left(-2,0\right)\cup \left(0,+\infty\right)$$
- step2: Multiply both sides of the equation by LCD:
  $$\left(\frac{1}{L}+\frac{1}{L+2}\right)\times 4L\left(L+2\right)=\frac{3}{4}\times 4L\left(L+2\right)$$
- step3: Simplify the equation:
  $$8L+8=3L^{2}+6L$$
- step4: Move the expression to the left side:
  $$8L+8-\left(3L^{2}+6L\right)=0$$
- step5: Calculate:
  $$2L+8-3L^{2}=0$$
- step6: Factor the expression:
  $$\left(2-L\right)\left(4+3L\right)=0$$
- step7: Separate into possible cases:
  $$\begin{align}&2-L=0\&4+3L=0\end{align}$$
- step8: Solve the equation:
  $$\begin{align}&L=2\&L=-\frac{4}{3}\end{align}$$
- step9: Check if the solution is in the defined range:
  $$\begin{align}&L=2\&L=-\frac{4}{3}\end{align},L \in \left(-\infty,-2\right)\cup \left(-2,0\right)\cup \left(0,+\infty\right)$$
- step10: Find the intersection:
  $$\begin{align}&L=2\&L=-\frac{4}{3}\end{align}$$
- step11: Rewrite:
  $$L_{1}=-\frac{4}{3},L_{2}=2$$
The solutions to the equation are $$ L = -\frac{4}{3} $$ and $$ L = 2 $$.

Since time cannot be negative, we discard the solution $$ L = -\frac{4}{3} $$.

Therefore, the time taken by the larger pipe to fill the tank is 2 hours, and the time taken by the smaller pipe to fill the tank is 2 + 2 = 4 hours.
The larger pipe takes 2 hours to fill the tank, and the smaller pipe takes 4 hours to fill the tank.

A water tank can be filled with water by two pipes
in 1 hour and 20 minutes. If the smaller pipe takes
2 hours longer than the larger pipe to fill the tank,
find the time taken by each pipe to fill the tank.

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