Respuesta :
We need to solve the system:
[tex]\begin{gathered} 3x=6y-4 \\ 4x+3y=-1 \end{gathered}[/tex]The first step is to isolate one of the variables and replace it at the other equation. We will isolate the "x" variable on the first equation:
[tex]x=\frac{6y}{3}-\frac{4}{3}[/tex]Then we replace this value on the second equation:
[tex]\begin{gathered} 4\cdot(\frac{6y}{3}-\frac{4}{3})+3y=-1 \\ \frac{24y}{3}-\frac{16}{3}+3y=-1 \\ \frac{24y-16+9y}{3}=-1 \\ 33y-16=-3 \\ 33y=-3+16 \\ 33y=13 \\ y=\frac{13}{33} \end{gathered}[/tex]We can use this value to determine x.
[tex]\begin{gathered} x=\frac{6\cdot\frac{13}{33}}{3}-\frac{4}{3} \\ x=\frac{\frac{78-132}{33}}{3}=\frac{\frac{-54}{33}}{3}=\frac{-54}{99}=\frac{-18}{33}=\frac{-6}{11} \end{gathered}[/tex]The solution to the system is (-6/11, 13/33)
