To verify if the answer satisfies the system of the equations, we can substitute the values of x, y, and z in the equations.
If the values are the same as the values in the equations, then x, y, and z are the solutions.
1) Let's substitute x, y, and z in the first equation.
x = 1
y = -1
z = 2
[tex]\begin{gathered} 4x+4y+2z=4 \\ 4\cdot1+4\cdot(-1)+2\cdot2=4 \\ 4-4+4=4 \\ 4=4 \\ \text{OK} \end{gathered}[/tex]
2) Let's substitute x, y, and z in the second equation.
[tex]\begin{gathered} x-y-z=0 \\ 1-(-1)-2=0 \\ 1+1-2=0 \\ 2-2=0 \\ 0=0 \\ OK \end{gathered}[/tex]
3) Let's substitute x, y, and z in the third equation.
[tex]\begin{gathered} 3y-3z=-8 \\ 3\cdot(-1)-3\cdot2=-8 \\ -3-6=-8 \\ -9=-8 \\ \text{NOT OK} \end{gathered}[/tex]
Since the values of x, y and z do not satisfy equation 3, it is not the solution of this system.
Answer: NO.
