Given the system:
[tex]\begin{gathered} 3x-9y=12\text{ Eq. 1} \\ 3x+4y=-1\text{ Eq. 2} \end{gathered}[/tex]
First, we solve for 3x on both equations, as follows:
[tex]\begin{gathered} 3x=12+9y \\ 3x=-1-4y \end{gathered}[/tex]
Equating both equations:
[tex]\begin{gathered} 12+9y=-1-4y \\ 13=-4-9y \\ 13=-13y \\ y=-1 \end{gathered}[/tex]
Substituting y on equation 2:
[tex]\begin{gathered} 3x+4y=-1\text{ Eq. 2.} \\ 3x+4\times(-1)=-1 \\ 3x=-1+4 \\ 3x=3 \\ x=1 \end{gathered}[/tex]
ANSWER
(1, -1)
