The first system of equations is
![\mleft\{\begin{aligned}5x+6y=-9 \\ 7x+2y=13\end{aligned}\mright.](https://img.qammunity.org/2023/formulas/mathematics/college/orw554052c3ah67x2rd52wgag1qg7x20m9.png)
To solve this system, we multiply the second equation by -3, that way we'll be able to eliminate y and solve for x.
![\begin{gathered} \mleft\{\begin{aligned}5x+6y=-9 \\ -21x-6y=-39\end{aligned}\mright. \\ 5x-21x+6y-6y=-9-39 \\ -16x=-48 \\ x=(-48)/(-16)=3 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/gx8j55w1qa55tif0vuv2nm9oi1v1nhnlct.png)
Then, we use the value of x to find y.
![\begin{gathered} 5x+6y=-9 \\ 5(3)+6y=-9 \\ 15+6y=-9 \\ 6y=-9-15 \\ 6y=-24 \\ y=(-24)/(6) \\ y=-4 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/p9la1p2kbvswlznhd8t55i1g5fyg9mxyar.png)
Therefore, the solutions are x = 3 and y = -4.
The second system of linear equations is
![\mleft\{\begin{aligned}2x-9y=-10 \\ 3x+5y=22\end{aligned}\mright.](https://img.qammunity.org/2023/formulas/mathematics/college/liip2plecqulrucuy6ugsrct25vci9fwp5.png)
To solve this system, we multiply the first equation by -3/2.
![\begin{gathered} \mleft\{\begin{aligned}-3x+(27)/(2)y=15 \\ 3x+5y=22\end{aligned}\mright. \\ -3x+3x+(27)/(2)y+5y=15+22 \\ (27y+10y)/(2)=37 \\ (37y)/(2)=37 \\ y=(37\cdot2)/(37) \\ y=2 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/749ivhyqh4aiehrlpcs4dg8f90cdfi91cw.png)
Then, we use this value to find x.
![\begin{gathered} 2x-9y=-10 \\ 2x-9(2)=-10 \\ 2x-18=-10 \\ 2x=18-10 \\ 2x=8 \\ x=(8)/(2) \\ x=4 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/i1bg0lww7r1tovjuew0a2b19nsel8mv2cj.png)
Therefore, the solutions are x = 4 and y = 2.