Hello,

You should know what you mean by "solving" the system of linear equations.

Since there are only 2 equations for 4 variables,

the space of solutions is 2-dimensional.

(If there were 2 equations for 2 variables, the solution space would often be 0-dimensional, yielding only one solution.)

In this case, the solution space is parametrized by 2 variables.

Assuming we take z and w for parameters,

you are asked to express (x, y, z, w) in terms of z and w.

R1 +R2 -> R1

{x + 2z + w = -3}

{-2x + y -3 z - 3w = 8}

R2 + 2R1 -> R2

{x +2z +w = -3}

{y + z - w = 2}

Thus,

(x, y, z, w)=(-2z-w-3, -z+w+2, z, w)

is the solution.

The solution space is

{(-3, 2, 0, 0) + z(-2, -1, 1, 0) + w(-1, 1, 0, 1)| z, w: arbitrary}.

Bye.