Given the system of equations, notice that we can divide by -2 the second equation to get the following equivalent equation:
![\begin{gathered} -(1)/(3)(3x+6y=-39) \\ \Rightarrow-x-2y=13 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/7yba7nwok7gwnqueazhifruze41puz63m1.png)
then, we woul have the following equivalent system of equations:
![\begin{cases}4x+2y=-{22} \\ -x-2y=13\end{cases}](https://img.qammunity.org/2023/formulas/mathematics/college/ly6rnbzgkga4b23ir3zzt7ujfpqi0ix144.png)
if we add both equations, we can find the value of x:
![\begin{gathered} 4x+2y=-22 \\ -x-2y=13 \\ ---------- \\ 3x=-22+13=-9 \\ \Rightarrow x=-(9)/(3)=-3 \\ x=-3 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/4wqbm6ror499a9nzugheiwo792r6l8b11z.png)
now that we know that x = -3, we can use this value to find the value of y:
![\begin{gathered} -(-3)-2y=13 \\ \Rightarrow-2y=13-3=10 \\ \Rightarrow y=(10)/(-2)=-5 \\ y=-5 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/xord53s0j9wktbdonnxbs26fc2hqju2kow.png)
therefore, the solutions are x = -3 and y = -5