Proving the irrationality of $\sqrt{5} = x \sqrt{7} + y$ I need to prove that there are no rational numbers $x, y$ that
$$\sqrt{5} = x \sqrt{7} + y$$
We know that square root of prime is irrational so $y = 5 - 7x$ so the only number for it to be rational is if $y = 0$ so $x = \frac{\sqrt{7}} {\sqrt {5}}$ but that is irrational or when $x = 0$, $y = \sqrt{5}$ which is irrational.
$x = \sqrt{5} - y \sqrt {7}$, so I'm stuck here.
Or I need to try to prove each at turn, like 
1 case : let $x =$ blah blah and do it as irrational prime proof
2 case : let $y =$ ???
Help would be appreciated. 

Assume we have rationals $x,y$ such that $\sqrt{5} = x\sqrt{7} + y$.
Square both sides, we get $5 = 7x^2 + y^2 + 2xy\sqrt{7}$.
So $\sqrt{7} = {{5 - 7x^2 - y^2} \over {2xy}}$ which is rational, contradiction.
