finding inverse of function in ordered pair notation $$f: \mathbb{R} \times \mathbb{R} \mapsto \mathbb{R} \times \mathbb{R} $$
where f is defined as $$f(x,y) =(\text{somethingforx},\text{somethingfory}) $$
I dont want to post the exact question because I would like to get it on my own, but I am having trouble finding the best way to find the inverse of a function when it is given in ordered pair notation such as here. Any methods on how to proceed would be very helpful. 
Thanks 
 A: The inverse relation of f is 
{ ( ( something for x, something for y), (x,y) ) | ((x,y) , (somethingforx, somethingfory) ) belong to f}. 
In general the inverse of R is 
{ (b,a) | (a,b) belong to R}
Here 
a = (x,y)
and
b = ( something for x, something for y)
A: In your example $f(x,y)=(5x-3y,7y-2x)$
you do it exactly the same way.
$f(w,u) = (5w - 3u, 7u-2w) = (x,y)$
Now solve for $(w,u)$ in terms of $(x,y)$.
$x = 5w -3u; y = 7u-2w$
$w = \frac {x+3u}5 = \frac {7u-y}2$
$2(x+3u) = 5(7u-y)$
$2x+5y = 29u$
$u = \frac {2x+5y}{29}$
And so $w =\frac {x+3 \frac {2x+5y}{29}}5 = \frac {7  \frac {2x+5y}{29} -y}2 =$
$\frac {35x + 15y}{145} = \frac {14x+6y}{58}=\frac {7x+3y}{29}$.
So $f^{-1}(x,y) = (w,u) = (\frac {7x+3y}{29},\frac {2x+5y}{29})$
But finding an inverse isn't the best way to prove something is invertible 
It is better to show if $f(x,y) = f(w,u)$ whether that must mean $(x, y) = (w,u)$ or not.
So if $(5x-3y, 7y-2x) = (5w-3u, 7u-2w)$ then
$5x - 3y = 5w - 3u$ so $5x - 5w = 3y-3u$ so $(x-w) = \frac 35(y-u)$
And $7y-2x = 7u-2w$ so $7y-7u = 2x-2w$ so $x-w = \frac 72(y-u)$.
So $\frac 35(y-u) = \frac 72(y-u)$ but that is only possible if $y-u=0$
So $y=u$ and so $x-w = 0$ and $x = w$.  So $(x,y) = (w,u)$ and $f$ is one to one.
