Given:
[tex]\begin{gathered} -6x-3y=18\ldots\ldots\ldots(1) \\ \frac{x^{2}}{9}+\frac{y^{2}}{36}=1\ldots\ldots\ldots\ldots(2) \end{gathered}[/tex]
Let us consider the equation (1),
[tex]\begin{gathered} -6x-3y=18 \\ -2x-y=6 \\ -y=2x+6 \\ y=-(2x+6)\ldots\ldots\ldots(3) \end{gathered}[/tex]
Substitute equation (3) in (2), we get
[tex]\begin{gathered} \frac{x^2}{9}+\frac{(-(2x+6))^2_{}}{36}=1 \\ \frac{4x^2}{36}+\frac{4x^2+36+24x}{36}=1 \\ \frac{4x^2+4x^2+36+24x}{36}=1 \\ 8x^2+36+24x=36 \\ 8x^2+24x=0 \\ 8x(x+3)=0 \\ x=0,x=-3 \end{gathered}[/tex]
Substitute x=0 and x=-3 in equation (3) we get,
[tex]\begin{gathered} y=-(2(0)+6) \\ =-6 \\ y=-(2(-3)+6) \\ =0 \end{gathered}[/tex]
Hence, the solutions are, (0,-6) and (-3,0).
Let us verify this, by substituting (0,-6) and (-3,0) in equation (1), we get
For (0, -6),
[tex]\begin{gathered} -6(0)-3(-6)=18 \\ 18=18 \end{gathered}[/tex]
For (-3, 0)
[tex]\begin{gathered} -6(-3)-3(0)=18 \\ 18=18 \end{gathered}[/tex]
Hence, it is verified.
