I am trying to prove that the function $f: \mathbb{R}^2 \to \mathbb{R}^2$ defined by $f(x,y) = (x+y, x + 2y)$ is bijective. Here is my attempt.
Suppose $f(m,n) = f(p,q)$ for $(m,n), (p,q) \in \mathbb{R}^2$. Then $(p + q, p + 2q) = (m + n, m + 2n)$, and we have $$ p + q = m + n, \qquad p + 2q = m + 2n. $$ But then $$ p + 2q = (p+q) + q = (m+n) + q = (m+n) + n, $$ and cancelling $m+n$ gives $q = n$. Substituting into the first equation, we get $m = p$. Therefore, $(m,n) = (p,q)$, so $f$ is injective.
Now, we prove that $f$ is surjective. Let $(a,b) \in \mathbb{R}^2$. We will find $u(a,b), v(a,b)$ such that $f(u,v) = (a,b)$. We have: $$ f(u,v) = (u+v, u + 2v) = (a,b), $$ so $u + v = a$ and $u + 2v = b$. In particular, $u = a - v$, and substituting into the second equation gives $$ b = u + 2v = (a-v) + 2v = v + a. $$ So $b = v + a$, and hence $v = b - a$. Then $$ u = a - v = a - (b-a) = 2a - b. $$ To check that these formulas work, we have: $$ f(2a - b, b - a) = ((2a - b) + (b-a), (2a-b) + 2(b-a)) = (a, b), $$ so $f$ is surjective, hence bijective.
How does this look?
Alternatively, you can use linear algebra in this question if you already know some basic theorems about it. Specifically, the function $f$ is a linear transformation that is represented by the matrix $A = \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}$. Thus the function is bijective iff $A$ is invertible. But it is since $\text{det} A = 1 \neq 0$ hence $A$ is invertible and $f$ is a bijection.