Solve the following system of equations:
4x − 2y − z = −5
x − 3y + 2z = 3
3x + y − 2z = −5
(0, 1, 3)
(0, −1, 3)
(0, 1, −3)
(0, −1, −3)

Multiplying both sides of the first equation by 2, we get that [tex]8x-4y-2z=-10[/tex]. Adding this to the second equation, we get that [tex]9x-7y=-7[/tex].

If we multiply both sides of the equation [tex]4x-2y=-2[/tex] by 7, we get [tex]28x-14y=-14[/tex].

If we multiply both sides of the equation [tex]9x-7y=-7[/tex] by 2, we get [tex]18x-14y=-14[/tex].

Subtracting the equation [tex]28x-14y=-14[/tex] from the equation [tex]18x-14y=-14[/tex], we get that [tex]-10x=0 \longrightarrow x=0[/tex].

This means that [tex]4(0)-2y=-2 \longrightarrow y=1[/tex]

And thus, [tex]4(0)-2(1)-z=-5 \longrightarrow -2-z=-5 \longrightarrow z=3[/tex].

So, the solution is (0, 1, 3).

Solve the following system of equations:4x − 2y − z = −5 x − 3y + 2z = 3 3x + y − 2z = −5(0, 1, 3)(0, −1, 3)(0, 1, −3)(0, −1, −3)​