PART A
To find the solutions to the system, we can multiply the second equation by -2
![\begin{cases}2y+x=12 \\ -2y=-(5)/(3)x+4\end{cases}](https://img.qammunity.org/2023/formulas/mathematics/college/gzkq33rp4398spy1u3m6pu6o8pmps8wt6j.png)
Then, we combine the equations and solve for y
![\begin{gathered} x=12-(5)/(3)x+4 \\ x+(5)/(3)x=16 \\ (3x+5x)/(3)=16 \\ 8x=16\cdot3 \\ x=(48)/(8) \\ x=6 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/etlzsgy78fydayngp435asuxysgkw0hyr2.png)
Now, we find y using x-value
![\begin{gathered} 2y+x=12 \\ 2y+6=12 \\ 2y=12-6 \\ y=(6)/(2) \\ y=3 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/yyy9nqwhx20djyjkf12fdjjpcq88fu6ovb.png)
Hence, the ordered pair where the graphs intersect is (6,3).
Part B.
To verify that the point (6,3) is a solution of 2y+x = 12, we just have to replace it
![\begin{gathered} 2\cdot3+6=12 \\ 6+6=12 \\ 12=12 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/7a52cauw8w02ik31336f4wwts6wvgxt6io.png)
As you can observe, it satisfies the equation, so it's a solution to it.
Part C.
To verify that the point (6,3) is a solution of the second equation, we just have to replace it
![\begin{gathered} y=(5)/(6)x-2 \\ 3=(5)/(6)\cdot6-2 \\ 3=5-2 \\ 3=3 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/q1f91h4dj7f14c9s5z89t98uwvvyxg34f4.png)
Hence, the point is a solution to the second equation too.
PART D.
Point (4,4) couldn't be a solution of the system because it has a unique solution, which is (6,3). Additionally, the point (4,4) only satisfies the first equation but not the second one, and the solution must be a solution to both equations.