Determining whether a multivariable function is onto A problem is asking me to identify whether the following function is onto: $$ R^2 -> R^2$$    $$f(x,y)=(x+y,2x+2y) $$
I understand the definition of onto (each real number solution in a certain domain being corresponded to) and it makes sense enough, when given ordinary linear equations like $$y = x^2 $$
which is not onto, considering no negative values of y can be reached. This can easily be seen graphically.  I'm more confused when given a function of more than one variable, such as the one above or 
$$R^3->R^3 $$
$$f(x,y,z)=(x+y,y+z,x+z) $$
Can I get some help putting these functions in similar terms to the y = 3x case? 
 A: For $f: R^2 \rightarrow R^2$, $f(x,y)=(x+y, 2x+2y)=(x+y,2(x+y))$, I claim that is it not onto. 
For example, we cannot find preimage of $f$ for $(0,1)$ since the second entry is not equal to twice of the first entry.
For $f: R^3 \rightarrow R^3$, $f(x,y, z)=(x+y,y+z,z+x)=(x,y,z)\begin{pmatrix} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{pmatrix}$
Let $A=\begin{pmatrix} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{pmatrix}$, the question is now given $(\alpha, \beta, \gamma)$, am I able to find $(x,y,z)$ such that $$(x,y,z)A=(\alpha, \beta, \gamma)$$
verify that $A$ is non-singular, and you can indeed compute
$$(x,y,z)=(\alpha, \beta, \gamma)A^{-1}$$
Hence, it is onto.
Edit:
$$x+y = \alpha$$
$$y+z = \beta$$
$$x+z = \gamma$$
where $\alpha, \beta, \gamma$ are given and we want to solve for $x,y,z$.
One possible way to solve for this problem is to first sum everything up and divides by $2$. We obtain
$$x+y+z=\frac{\alpha+\beta+\gamma}{2}$$
$$(x+y+z)-(x+y)=z=\frac{\alpha+\beta+\gamma}{2}-\alpha$$
Similarly, we can solve for $x$ and $y$
