Answer:
x=2,
z=-4,
y=3
Step-by-step explanation:
[tex]\begin{bmatrix}5x-2y+z=0\\ 2x-y+z=-3\\ 3x+4y=18\end{bmatrix}\\\\\mathrm{Isolate}\:x\:\mathrm{for}\:3x+4y=18:\quad x=\frac{18-4y}{3}\\\\\mathrm{Subsititute\:}x=\frac{18-4y}{3}\\\\\begin{bmatrix}5\times\frac{18-4y}{3}-2y+z=0\\ \\2\times\frac{18-4y}{3}-y+z=-3\end{bmatrix}\\\\Simplify\\\\\begin{bmatrix}z+\frac{90-26y}{3}=0\\ \\z+\frac{36-11y}{3}=-3\end{bmatrix}\\\\\mathrm{Isolate}\:z\:\mathrm{for}\:z+\frac{90-26y}{3}=0:\\\\\quad z=-\frac{-26y+90}{3}\\\\\mathrm{Subsititute\:}z=-\frac{-26y+90}{3}[/tex]
[tex]\begin{bmatrix}-\frac{-26y+90}{3}+\frac{36-11y}{3}=-3\end{bmatrix}\\\\Simplify\\\begin{bmatrix}5y-18=-3\end{bmatrix}\\\\\mathrm{Isolate}\:y\:\mathrm{for}\:5y-18=-3:\quad y=3\\\\\mathrm{For\:}z=-\frac{-26y+90}{3}\\\\\mathrm{Subsititute\:}y=3\\\\z=-\frac{-26\times\:3+90}{3}\\\\-\frac{-26\times\:3+90}{3}=-4\\\\z=-4\\\\\mathrm{For\:}x=\frac{18-4y}{3}\\\\\mathrm{Subsititute\:}z=-4,\:y=3\\\\x=\frac{18-4\times\:3}{3}\\\\\frac{18-4\times\:3}{3}=2\\\\x=2\\\\x=2,\:\\z=-4,\:\\y=3[/tex]
