How can I solve $4x + 51y = 9$ using congruences? I'm given: $4x+51y=9$.

I am given a hint that when we use $4x=9 \pmod{51}$ we get $x = 15 + 15t$, and also if we use the congruence $51y=9 \pmod 4$ we get $y=3+4s$. They say it's handy to then find the relation between $s$ and $t$.

I have no idea how they got those suggestions and I need to know how to do that too.

I'm really stuck :(
Write: $4x + 51y = 9$ as $9x - 5x + 45y + 6y = 9$ $\to$ $9| (-5x + 6y)$ $\to$ $-5x + 6y = 9h$ $\to$ 

$-5x = 3(3h - 2y)$ $\to$ $3|(-5x)$ $\to$ $3|x$.  $x = 3k$. Back to the main 

equation: $4(3k) + 51y = 9$ $\to$ $51y = 9 - 12k$ $\to$ $17y = 3 - 4k$ $\to$ $y = \dfrac{3 - 4k}{17}$. In order 

for $y$ to be an integer we must have: $k = 5 + 17t$. Thus $x = 3k = 3(5 + 17t) = 15 + 51t$. So

$y = \dfrac{3 - 4(5 + 17t)}{17} = \dfrac{-17 - 68t}{17} = -1 - 4t$.

Thus the solution is: $(x,y) = \{(15 + 51t, -1 - 4t): t \in \mathbb{Z}\}$.

Check: $4x + 51y = 4(15 + 51t) + 51(-1 - 4t) = 60 + 204t - 51 - 204t = 9$.
modulo 51.
4x=9(mod51)
51=3×17
4x=9(mod3) then x=0(mod3) then x=3k k:integer
4x=9(mod17) then 12k=9(mod17)
9k=60k=45(mod17)
k=5(mod17) then k=17t+5 t:integer
x=3k=3(17t+5)=51t+15

modulo4.
51y=9(mod4) then -y=1(mod4)
-y=4t+1 then y=-(4t+1)=-4t-1