Answer:
[tex]2x+3y+4z=1;3y+3z=6;x=-3z[/tex]Explanation:
We're given the solution to a system of equations as (-3, 1, 1), that is, x = -3, y = 1, and z = 1.
So to determine which of the given system of equations has the above solution, we'll substitute the given values of x, y, and z into each of the equations, and whichever one gives the correct result will be the right answer.
For the 1st option, let's consider the first equation in the system of equations;
[tex]\begin{gathered} x-y-z=-2 \\ -3-1-1\ne-2 \\ -5\ne-2 \end{gathered}[/tex]We can see that the first option is not the right answer since the RHS and LHS are not equal to each other in one of the equations.
For the 2nd option, let's consider the 1st equation;
[tex]\begin{gathered} x-y+3z=2 \\ -3-1+3(1)=2 \\ -3-1+3=2 \\ -4+3=2 \\ -1\ne2 \end{gathered}[/tex]We can also see that the 2nd option is not the correct answer.
For the 3rd option, let's pick the 1st equation;
[tex]\begin{gathered} 4x+y-3z=-4 \\ 4(-3)+1-3(1)=-4 \\ -12+1-3=-4 \\ -14\ne-4 \end{gathered}[/tex]We can see also that the 3r option isn't the correct answer.
For the 4th option, let's pick the 1st equation;
[tex]\begin{gathered} 2x+3y+4z=1 \\ 2(-3)+3(1)+4(1)=1 \\ -6+3+4=1 \\ 1=1 \end{gathered}[/tex]Let's pick the 2nd equation;
[tex]\begin{gathered} 3y+3z=6 \\ 3(1)+3(1)=6 \\ 3+3=6 \\ 6=6 \end{gathered}[/tex]Let's pick the 3rd equation;;
[tex]\begin{gathered} x=-3z \\ -3=-3(1) \\ -3=-3 \end{gathered}[/tex]Since the solution (-3, 1, 1) satisfies the system of equations in the 4th option as seen above, then the 4th option is the right answer.
