According to the information given in the exercise:
- The empty tank is filled in 10 hours.
- The variable "x" represents the time (in hours) it takes pipe A to fill the tank and "y" represents the time (in hours) it takes pipe B to fill the tank.
- Pipe used A alone is used for 6 hours and then it is turned off.
- Pipe B finish filling in 18 hours (after pipe A is turned off).
By definition, these formulas can be used for Work-Rate problems:
![\begin{gathered} (t)/(t_1)+(t)/(t_2)=1 \\ \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/lwnmftqeh4m7nmm7prkywi8boaicl1shnb.png)
![(1)/(t_1)+(1)/(t_2)=(1)/(t)](https://img.qammunity.org/2023/formulas/mathematics/college/q99ylhhy9r8qbnzdrqj0xqcztc8fk5540s.png)
Where:
- This is the individual time for the first object:
![t_1](https://img.qammunity.org/2023/formulas/mathematics/high-school/6t8raerqwwwndod9lzlgdmr7lqpvq16vt4.png)
-This is the individual time for the second object:
![t_2](https://img.qammunity.org/2023/formulas/physics/college/zqq3q7te6zwjc185awuay2537z7n9oy8r1.png)
- And "t" is the time for both objects together.
In this case, having the first equation:
![(1)/(x)+(1)/(y)=(1)/(10)](https://img.qammunity.org/2023/formulas/mathematics/college/xzu5lrtvi8ovvexbggxogpc8pwcxhou5xn.png)
You can set up the second equation:
![(6)/(x)+(18)/(y)=1](https://img.qammunity.org/2023/formulas/mathematics/college/t2mpkdu0aqmunvk2sxywcgsq38uthzjtdu.png)
Notice that the sum of that fraction is equal to the part of the tank filled: 1 (the whole tank).
Now you can set up the System of equations:
![\begin{cases}(1)/(x)+(1)/(y)=(1)/(10) \\ \\ (6)/(x)+(18)/(y)=1\end{cases}](https://img.qammunity.org/2023/formulas/mathematics/college/gy5szfgctjlu1r7le3sgj25kpxuel7mmku.png)
To solve it, you can apply the Elimination Method:
1. Multiply the first equation by -6.
2. Add the equations.
3. Solve for "y".
Then:
![\begin{cases}-(6)/(x)-(6)/(y)=-(6)/(10) \\ \\ (6)/(x)+(18)/(y)=1\end{cases}](https://img.qammunity.org/2023/formulas/mathematics/college/5shq0a8pjfcvflb7vox9e7qgz1y4jnz0w3.png)
![\begin{gathered} \begin{cases}-(6)/(x)-(6)/(y)=-(6)/(10) \\ \\ (6)/(x)+(18)/(y)=1\end{cases} \\ ------------- \\ 0+(12)/(y)=(2)/(5) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/bi1rpia44fn1vjdehz8spv1sdsyakk83xa.png)
![\begin{gathered} 12=(2)/(5)y \\ \\ 12\cdot5=2y \\ \\ (60)/(2)=y \\ \\ y=30 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/foj7ttr0t2z3pt207vnm63k8g4rpg98xkh.png)
4. Substitute the value of "y" into one of the original equations.
5. Solve for "x".
Then:
![\begin{gathered} (1)/(x)+(1)/(30)=(1)/(10) \\ \\ (1)/(x)=(1)/(10)-(1)/(30) \\ \\ (1)/(x)=(1)/(15) \\ \\ (15)(1)=(1)(x) \\ x=15 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/2x2vryp207s32a8hs8nl62kou3nuihrctz.png)
Therefore, the answer is:
- It will take pipe A 15 hours to fill the tank alone.
- It will take pipe B 30 hours to fill the tank alone.