We have to solve the system:
![\begin{gathered} -(5)/(7)-(11)/(7)x=-y \\ 2y=7+5x \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/7oxy62e9fkkkixvd0won2qwiw53mnuujy7.png)
To solve this system by elimination we have to add or substract a linear combination of the second equation from the first equation in order to eliminate one of the variables.
In this case we can multiply the first equation by 2 and add it to the second equation:
![\begin{gathered} -y\cdot2=(-(5)/(7)-(11)/(7)x)\cdot2 \\ -2y=-(10)/(7)-(22)/(7)x \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/h00qp422xjfx22a8vaqh7715ujmhig73xt.png)
![\begin{gathered} -2y+2y=(-(10)/(7)-(22)/(7)x)+(7+5x) \\ 0=-(10)/(7)-(22)/(7)x+7+5x \\ (22)/(7)x-5x=7-(10)/(7) \\ (22x-5\cdot7x)/(7)=(7\cdot7-10)/(7) \\ 22x-35x=49-10 \\ -13x=39 \\ x=(39)/(-13) \\ x=-3 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/xdo9h787uwy2s9h6g2no04z6xrie32yeah.png)
Now we can use any of the two equations to find y:
![\begin{gathered} -y=-(5)/(7)-(11)/(7)x \\ y=(5)/(7)+(11)/(7)(-3) \\ y=(5)/(7)-(33)/(7) \\ y=-(28)/(7) \\ y=-4 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/doj2pthbo64l9dea4tgpdx6acon9o34sun.png)
Answer: x=-3 and y=-4