Possibilty of a number $n$ to be a multiple of $x$ that has a remainder of $y$ Is there a formula or solution for this :
"$n$ is a number that is a multiple of $x$ and when divided by $b$ has remainder of $y$".
Is there a fast way to finding this number?

The short answer is "No". Here is the long answer.
$n \equiv 0 \pmod x \implies n = \alpha x \ \text{for some integer $\alpha$}$
$n \equiv y \pmod b \implies \alpha x \equiv y \pmod b$
Let $g = \gcd(x, b)$. Then there is no solution unless $g \mid y$.
If $g \mid y$...
Let $X = \dfrac xg, \quad Y = \dfrac yg, \quad  B = \dfrac bg.$ 
Then $\alpha X \equiv Y \pmod B$ and $\gcd(X,B) = 1$.
So there exists integers $\xi, \beta$ such that $\xi X + \beta B = 1$.
Then $\xi Y \equiv \alpha \xi X \equiv \alpha - \alpha \beta B \equiv \alpha \pmod B$.
That is $\alpha \equiv \xi Y \pmod B$
Example.
Find a number $n$ that is a multiple of $20$ and leaves a remainder of $24$ when divided by $34$.
We have $x = 20, \quad y=24, \quad b=34$.
$n \equiv 0 \pmod{20} \implies n = 20 \alpha \ \text{for some integer $\alpha$}$.
$n \equiv 24 \pmod{34} \implies 20 \alpha \equiv 24 \pmod{34}$
Let $g = \gcd(20, 34) = 2$. Then there is no solution unless $2 \mid 24$. Which it does.
Let $X = \dfrac{20}{2} = 10, \quad Y = \dfrac{24}{2}=12, \quad  B = \dfrac{34}{2}=17.$ 
Then $10 \alpha \equiv 12 \pmod{17}$ and $\gcd(10,17) = 1$.
So there exists integers $\xi, \beta$ such that $10 \xi + 17 \beta = 1$.
\begin{array}{r|rrr|l}
    & 17 &  1 & 0  & 17 = (1)17 + (0)10\\
 -1 & 10 &  0 & 1  & 10 = (0)17 + (1)10\\
\hline
 -1 &  7 &  1 & -1 &  7 = (1)17 + (-1)10\\
 -2 &  3 & -1 &  2 &  3 = (-1)17 + (2)10 \\
\hline
    &  1 &  3 & -5 &  1 = (3)17 + (-5)10
\end{array}
We find $\xi = 3, \quad \beta = -5$
So $\alpha \equiv -5 \cdot 12 = -60 \pmod{17} = 8 \pmod{17}$
and $n = 20\alpha = 160$
CHECK:
$160$ is a multiple of $20$
$160 = 4 \cdot 34 + 24$
