Consider $$f: \mathbb{R} \times \mathbb{R} \to \mathbb{R} \times \mathbb{R}, \, (x, \,y) \mapsto (x+2y, \, 2x-y).$$

I know that $f$ is bijective, so it's also injective and it has a left inverse $l$, which means that $l \circ f = Id\_{\mathbb{R} \times \mathbb{R}}$.

For any $(x, y) \in \mathbb{R} \times \mathbb{R}$, $l(f(x,y))=(x,y)$ has to hold, so I thought I could determine $l$ like this:

$l(f(x,y))=(x,y) \Leftrightarrow l(x+2y, 2x-y) = (x,y)$.

This yields the linear system of equations $$\begin{align} x&+2y=x \\ 2x&-y=y \end{align}$$ which gives $x=y=0$, but I haven't learned anything about $l$.

I already know that $l$ is defined by $l: \mathbb{R} \times \mathbb{R} \to \mathbb{R} \times \mathbb{R}, \, (x,\,y) \mapsto \left(\dfrac{2y+x}{5},\,\dfrac{2x-y}{5}\right)$, but why does my approach completely fail?