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**Hint:**

Here, we will assume the time taken by both the pipes to fill the swimming pool to be some variable. We will frame the linear equations of two variables using the given information. Then we will solve the equations using the elimination method to find the values of the variable and find the time taken by both the pipes to fill the swimming pool.

**Complete step by step solution:**

Let the two pipes used to fill the swimming pool be \[A\] and \[B\]. We are given that Pipe \[A\] has larger diameter than Pipe \[B\].

Now, let us consider that Pipe \[A\] take \[x\] hours and Pipe \[B\] take \[y\] hours to fill the swimming pool separately.

Pipe \[A\] can fill the \[\dfrac{1}{x}\] of the swimming pool in 1 hour.

Pipe \[B\] can fill the \[\dfrac{1}{y}\] of the swimming pool in 1 hour.

If Pipe \[A\] and Pipe \[B\]are opened simultaneously, then they take 12 hours to fill the tank.

If they are together, then in 1 hour, they will fill the pool:

\[\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{{12}}\] ……………………………………….. \[\left( 1 \right)\]

Now, In 4 hour Pipe \[A\] can fill the pool. Thus we get \[\dfrac{4}{x}\].

Now, In 9 hour Pipe \[B\] can fill the pool. Thus we get \[\dfrac{9}{y}\].

They will fill half of the pool. Thus we get

\[\dfrac{4}{x} + \dfrac{9}{y} = \dfrac{1}{2}\]……………………………………………… \[\left( 2 \right)\]

Let \[\dfrac{1}{x} = a\] and \[\dfrac{1}{y} = b\], so substituting these in the equations \[\left( 1 \right)\], we get

\[a + b = \dfrac{1}{{12}}\]

On cross multiplication, we get

\[ \Rightarrow 12a + 12b = 1\]………………………………………….. \[\left( 3 \right)\]

Again substituting \[\dfrac{1}{x} = a\] and \[\dfrac{1}{y} = b\] in the equations \[\left( 2 \right)\], we get

\[4a + 9b = \dfrac{1}{2}\]

On cross multiplication, we get

\[ \Rightarrow 8a + 18b = 1\] …………………………………………………….. \[\left( 4 \right)\]

Multiplying equation \[\left( 3 \right)\] by 2, we get

\[24a + 24b = 2\]……………………………….. \[\left( 5 \right)\]

Multiplying equation \[\left( 4 \right)\] by 3, we get

\[24a + 54b = 3\]…………………………. \[\left( 6 \right)\]

Subtracting equation \[\left( 5 \right)\] from \[\left( 6 \right)\], we get

\[24a + 54b - \left( {24a + 24b} \right) = 3 - 2\]

\[ \Rightarrow 24a + 54b - 24a - 24b = 1\]

Subtracting the like terms, we get

\[ \Rightarrow - 30b = - 1\]

\[ \Rightarrow b = \dfrac{1}{{30}}\]

Substituting the value of \[b\] in equation \[\left( 3 \right)\], we get

\[ \Rightarrow 12a + 12\left( {\dfrac{1}{{30}}} \right) = 1\]

By simplifying and rewriting the equation, we get

\[ \Rightarrow 12a = 1 - \dfrac{{12}}{{30}}\]

Taking LCM on the RHS, we get

\[ \Rightarrow 12a = 1 \times \dfrac{{30}}{{30}} - \dfrac{{12}}{{30}}\]

\[ \Rightarrow a = \dfrac{{18}}{{30 \times 12}}\]

Simplifying the expression, we get

\[ \Rightarrow a = \dfrac{1}{{20}}\]

Now substituting \[a = \dfrac{1}{{20}}\] in \[\dfrac{1}{x} = a\], we get

\[x = \dfrac{1}{a} = 20\]

Substituting \[b = \dfrac{1}{{30}}\] in \[\dfrac{1}{y} = b\] , we get

\[y = \dfrac{1}{b} = 30\]

**Thus, Pipe \[A\] takes 20 hours and pipe \[B\] takes 30 hours to fill the swimming pool.**

**Note:**

Here, we have formed linear equations based on the given condition. A linear equation is an equation that has the highest degree of variables as 1. A linear equation has only one solution.

Here we have multiplied equation \[\left( 3 \right)\] by 2 and equation \[\left( 4 \right)\] by 3 because we want the coefficient of either variable to be equal so that we can eliminate that variable and find the second variable easily. This method is called the elimination method because we eliminate one variable to find the other.

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