Find all first degree functions that satisfy... 
I want to find all first degree functions that satisfy:
  $$f(x,y) = f(x+1, y-1) - 5 = f(x-2, y+1) + 8$$

I know that a first degree function looks like $$f(x,y) = ax+by+c$$
Please point me in the right direction.
Thank you in advance.
 A: HINT: $$f(x+1,y-1)-5=f(x-2,y+1)+8$$ means
$$a(x+1)+b(y-1)+c-5=a(x-2)+b(y+1)+c+8$$
can you go further?
from here we get
$$3a-2b=13$$
A: You have
$$
\eqalign{
  & \left\{ \matrix{
  f(x,y) = f(x + 1,y - 1) - 5 \hfill \cr 
  f(x,y) = f(x - 2,y + 1) + 8 \hfill \cr}  \right.  \cr 
  & \left\{ \matrix{
  ax + by + c = ax + a + by - b - 5 \hfill \cr 
  ax + by + c = ax - 2a + by + b + 8 \hfill \cr}  \right.  \cr 
  & \left\{ \matrix{
  c = a - b - 5 \hfill \cr 
  c =  - 2a + b + 8 \hfill \cr}  \right.  \cr 
  & \left\{ \matrix{
  2c =  - a + 3 \hfill \cr 
  0 = 3a - 2b - 13 \hfill \cr}  \right. \cr} 
$$
and since they are two equations in three unknowns
you can solve for any two unknown in function of the remaining one.
A: Since equations must hold for all $x,y$ we can put for example $x=0$ and $y=0$ (you culd put somthing else also). Then we get:
$$ f(0,0) = f(1,-1)-5 = f(-2,1)+8$$ thus we have
$$ c=a-b+c-5 = -2a+b+c+8$$
From $c=a-b+c-5$ we get $\boxed{a-b=5}\;\;\;\;\;\; (1)$ 
and 
from $c=-2a+b+c +8$ we get $\boxed{-2a+b=-8} \;\;\;\;\;\; (2)$
Now if you add (boxed) equations $(1)$ and $(2)$ you get $-a=-3$ thus $a=3$ and $b=-2$. Yet we don't have any other limitation for $c$, so $c$ can be any number.
So solution of your problem is(are) function(s):
$$ f(x,y)=3x-2y+c,\;\;\;\;\;\;  c\in \mathbb{R}$$
You can easly check that this function(s) satisfyes starting equations. 
