Answer:
correct choice is A
Step-by-step explanation:
From the second equation express x and substitute it into the first and third equations:
[tex]\left\{\begin{array}{l}x=4-2z\\5(4-2z)+2y+z=4\\2(4-2z)+y-z=-1\end{array}\right..[/tex]
Then
[tex]\left\{\begin{array}{l}x=4-2z\\20-10z+2y+z=4\\8-4z+y-z=-1\end{array}\right.\Rightarrow\left\{\begin{array}{l}x=4-2z\\2y-9z=-16\\y-5z=-9\end{array}\right..[/tex]
From the third equation [tex]y=-9+5z[/tex] and substituting it into the second, you get
[tex]2(-9+5z)-9z=-16,\\ \\-18+10z-9z=-16,\\ \\z=-16+18,\\ \\z=2.[/tex]
If [tex]z=2,[/tex] then
[tex]y=-9+5\cdot 2=1[/tex]
and
[tex]x=4-2\cdot 2=0.[/tex]
The solution of the system is [tex]x=0,\ y=1,\ z=2.[/tex]
