Answer:
Please check the explanation below.
Step-by-step explanation:
Given the equations
[tex]6x+18y+6z=24,\:-x-3y-2z=-4,\:2x+6y+2z=8\:\:[/tex]
Steps to solve
[tex]\begin{bmatrix}6x+18y+6z=24\\ -x-3y-2z=-4\\ 2x+6y+2z=8\end{bmatrix}[/tex]
[tex]\mathrm{Subsititute\:}x=4-3y-z[/tex]
[tex]\begin{bmatrix}-\left(4-3y-z\right)-3y-2z=-4\\ 2\left(4-3y-z\right)+6y+2z=8\end{bmatrix}[/tex]
As
[tex]-\left(4-3y-z\right)-3y-2z=-4[/tex]
[tex]-z-4=-4[/tex]
and
[tex]2\left(4-3y-z\right)+6y+2z=8[/tex]
[tex]8=8[/tex]
so
[tex]\begin{bmatrix}-z-4=-4\\ 8=8\end{bmatrix}[/tex]
Isolating z for [tex]-z-4=-4[/tex]
[tex]-z-4=-4[/tex]
[tex]-z=0[/tex]
[tex]\frac{-z}{-1}=\frac{0}{-1}[/tex]
[tex]z=0[/tex]
[tex]\mathrm{For\:}x=4-3y-z[/tex]
[tex]\mathrm{Expressing\:}x\mathrm{\:in\:terms\:of\:}y[/tex]
[tex]x=4-3y-0[/tex]
[tex]x=-3y+4[/tex]
Therefore,
[tex]\mathrm{The\:solutions\:to\:the\:system\:of\:equations\:are:}[/tex]
[tex]x=-3y+4,\:z=0[/tex]
