A system of linear equations $\begin{cases}x+2y-z=3,\\z+w=1;\end{cases}$

Question

So I've been given this system of equations $$\begin{cases} x + 2y - z = 3,\\ z + w = 1; \end{cases}$$ and I had to find the solutions.

I got stuck up to

$$\begin{cases} x+2y-(1-w)=3,\\ x+2y+w=4; \end{cases}$$

I don't know what to do from here.

Answer 1

We have $z=x+2y-3$ and $w=1-z=-x-2y+4$.

So if $x,y\in\mathbb{R}$ then $(x,y,x+2y-3,-x-2y+4)$ is a solution.

Now if $(x,y,z,w)$ is a solution it can be written as $(x,y,x+2y-3,-x-2y+4)$.

Therefore the set of solutions of this system in $\mathbb{R}^4$ is:$$S=\{(x,y,x+2y-3,-x-2y+4)\,:\,x,y\in\mathbb{R}\}$$

With more linear algebra, notice that:$\forall x,y\in\mathbb{R},\,(x,y,x+2y-3,-x-2y+4)=(x,0,x,-x)+(0,y,2y,-2y)+(0,0,-3,4)=x(1,0,1,-1)+y(0,1,2,-2)+(0,0,-3,4)$.

Thus $$S=(0,0,-3,4)+\text{span}\{(1,0,1,-1),(0,1,2,-2)\}$$