We have the following equations:
First equation
[tex]x-y=3[/tex]Second equation
[tex]\frac{1}{2}x+y=3[/tex]The first thing we will do is clear the first equation:
[tex]\begin{gathered} x-y=3 \\ x=3+y \end{gathered}[/tex]The second thing we will do is replace the found value of "x" in terms of "y", in the second equation and solve for "y"
[tex]\begin{gathered} \frac{1}{2}x+y=3 \\ \frac{1}{2}(3+y)+y=3 \\ \frac{3}{2}+\frac{y}{2}+y=3 \\ \frac{3}{2}y=3-\frac{3}{2} \\ \frac{3}{2}y=\frac{3}{2} \\ y=\frac{3}{2}\cdot\frac{2}{3} \\ y=1 \end{gathered}[/tex]In conclusion, the value of "y" is 1
The third thing we will do is to replace the found value of "y" in the first equation
[tex]\begin{gathered} x=3+1 \\ x=4 \end{gathered}[/tex]In conclusion, the solution to this system of equations is:
[tex]\begin{gathered} y=1 \\ x=4 \\ (4,1) \end{gathered}[/tex]Therefore, (6,3) is not the solution to this system of equations.