This problem can be seen as the next equation
![\begin{gathered} 4F+2H+1O=T \\ F\text{ = dollars spent on flights} \\ H\text{ = dollars spent at hotels} \\ O\text{ = dollars spent in other stuff} \\ T\text{ = total of points} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/f3807iq6kf0obr0a7q30mtdu4e8e9vayw9.png)
Since he have charged 9480 dollars in total then we also have the next equation.
![F+H+O=9480](https://img.qammunity.org/2023/formulas/mathematics/college/xenkvbbpa4bvv6fvpuw4evsp16mor850qy.png)
With the rest of the information, we conclude the next system of equations
![\begin{gathered} 4F+2H+O=14660 \\ F+H+O=9480 \\ F-2H=140 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/3el163fbdddgyctsb87waoqa07ogp9qh5n.png)
From the first two equations, we can substract them to obtain
![\begin{gathered} 4F+2H+O=14660 \\ - \\ F+H+O=9480 \\ = \\ 3F+H=5180 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/psglyzkasp972qtiaem1q9dxnwlhk8kw6h.png)
If we multiply this last equation by 2 and add it to the last equation in the system we have
![\begin{gathered} 6F+2H=10360 \\ + \\ F-2H=140 \\ = \\ 7F\text{ = }10500 \\ \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/i6xwr7qd9jcymef6u8amyysn9m4wisuekb.png)
So
![\begin{gathered} F=(10500)/(7)=1500 \\ H=(1500-140)/(2)=680 \\ O=9480-1500-680=7300 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/91i3eqex3b68dfxgi8lvtpkhk46xw7sgpc.png)
Then, he spent 1500 dlls. on flights, 680 dlls. at hotels and 7300 dlls. on others
![\begin{gathered} 4\text{ 2 1 14660 } \\ 1\text{ 1 1 9480} \\ 1\text{ -2 0 140 }\approx \\ 4\text{ 2 1 14660 } \\ 3\text{ 1 0 5180} \\ 1\text{ -2 0 140 }\approx \\ 4\text{ 2 1 14660 } \\ 3\text{ 1 0 5180} \\ 7\text{ 0 0 }10500 \\ \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/1jj1m4seydzsrt9vnq2zv0hw1zfwdrt6ap.png)