Answer:
The solution of the system of equations will be:
[tex]y=-12,\:x=-3[/tex]
Step-by-step explanation:
Given the system of equations:
[tex]\begin{bmatrix}y-2x=-6\\ y-4x=0\end{bmatrix}[/tex]
Isolate [tex]y[/tex] for [tex]y-2x=-6[/tex], [tex]y=-6+2x[/tex]
[tex]\mathrm{Subsititute\:}y=-6+2x[/tex]
[tex]\begin{bmatrix}-6+2x-4x=0\end{bmatrix}[/tex]
[tex]\begin{bmatrix}-6-2x=0\end{bmatrix}[/tex]
Isolate x for [tex]-6-2x=0[/tex], [tex]x=-3[/tex]
[tex]\mathrm{For\:}y=-6+2x[/tex]
[tex]\mathrm{Subsititute\:}x=-3[/tex]
[tex]y=-6+2\left(-3\right)[/tex]
[tex]y=-12[/tex]
Therefore, the solution of the system of equations will be:
[tex]y=-12,\:x=-3[/tex]
