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?
The wrong step is when you go from $l(x+2y,2x-y) = (x,y)$ to $x+2y=x$ and $2x-y=y$. That's just not justified.
You know $f$ is surjective; so there is a value of $(x,y)$ that maps to $(3,5)$. So, we want to find $(x,y)$ such that $f(x,y)=(3,5)$. Using your step above, we would say "Oh, that means $x=3$ and $y=5$." Which of course is wrong. Instead, we need to evaluate $f(x,y)$ to $(x+2y,2x-y)$, and then set that equal to $(3,5)$, leading to $(x+2y,2x-y)=(3,5)$ and the equations $x+2y=3$ and $2x-y=5$.
So from $l(x+2y,2x-y)=(x,y)$, you would first need to replace the expression $l(x+2y,2x-y)$ by whatever its value is, then set that to $(x,y)$. But you don't know what $l(x+2y,2x-y)$ is in the first place! So you cannot do that.
You could do something as follows: you know that the function $l$ is linear, so there must exist constants $a,b,c,d$ such that $l(r,s) = (ar+bs, cr+ds)$. We just need to figure out $a,b,c,d$. So from $$l(x+2y,2x-y)=(x,y)$$ we get $$\Bigl( a(x+2y) + b(2x-y), c(x+2y) + d(2x-y)\Bigr) = (x,y)$$ and solve for $a,b,c,d$.
However, that just amounts to finding the inverse of the usual matrix.