Respuesta :
We are given the following two equations
[tex]\begin{gathered} 3x+y=-1\qquad eq.1 \\ 4x+3y=12\qquad eq.2 \end{gathered}[/tex]We are asked to solve the equation using the substitution method.
Let us re-arrange the eq.1 for y and then substitute it into the eq.2
[tex]\begin{gathered} 3x+y=-1 \\ y=-1-3x \end{gathered}[/tex]Now substitute it into the eq.2
[tex]\begin{gathered} 4x+3y=12 \\ 4x+3(-1-3x)=12 \\ 4x-3-9x=12 \\ -5x-3=12 \\ -5x=12+3 \\ -5x=15 \\ x=\frac{15}{-5} \\ x=-3 \end{gathered}[/tex]Now substitute the value of x into the eq.1
[tex]\begin{gathered} 3x+y=-1 \\ 3(-3)+y=-1 \\ -9+y=-1 \\ y=-1+9 \\ y=8 \end{gathered}[/tex]Therefore, the solution of the given equations is (-3, 8)
x = -3
y = 8
