Proof by induction for systems of equations with arithmetic progression A linear system of equations of the form
$\begin{cases} ux+(u+d)y=u+2d\\
ax+(a+i)y=a+2i\end{cases}$
Will always have the solution $x=-1, y=2$ (easily proven through matrix algebra). How can I prove this by induction?
 A: We can prove this using the induction scheme that says it's true for all $u,d,a,i$ if it's true for $u=d=a=i=1$ and also the (four) induction step(s) given it is true for $(u,d,a,i)$ then it also holds for $(u+1,d,a,i)$,  $(u,d+1,a,i)$, $(u,d,a+1,i)$ and $(u,d,a,i+1)$.

The base case will be $a,u,d,i = 1$ $$\begin{cases} x+2y=3\\
x+2y=3\end{cases}$$ clearly the solution holds for this case.

In all four cases the induction hypothesis will be $$\begin{cases} ux+(u+d)y=u+2d\\
ax+(a+i)y=a+2i\end{cases}$$

I will only show one induction step because the other three are so similar
We want to prove
$$\begin{cases} (u+1)x+(u+1+d)y=u+1+2d\\
ax+(a+i)y=a+2i\end{cases}$$
by algebra that's the same as
$$\begin{cases} ux+(u+d)y+x+y=u+2d+1\\
ax+(a+i)y=a+2i\end{cases}$$
and since $x+y = 1$ this reduces to our hypothesis.
A: As proposed by Bill Dubuque, a simple proof can be formulated like this 
We substitue $x=-1$ and $y=2$ into $ux+(u+d)y=u+2d$
$\Rightarrow -u+2(u+d)=u+2d$
To prove the above we must show that for $P(u,d)$, every of the following is true: $P(0,0)$, $P(u+1,d)$ and $P(u,d+1)$.
