<s>[INST] An integer when divided by $5$ and $13$ leaves residues $4$ and $7$ respectively Find an integer when divided by $5$ and $13$ leaves residues $4$ and $7$ respectively. (Without Modular Arithmetic).
I don't know if it is right, but I got this
$$n=5x+4=13y+7$$
$$n=5(x-11)+59=13(y-4)+59$$
The $\operatorname{lcd}(5,13)=65$
So our number is every number like $n=65k+59$
I'll be glad if you enlighten me it. 
 [/INST] $5x - 13y = 7-4 = 3 \Rightarrow 5x = 15y + 3 - 2y \Rightarrow x = 3y + \dfrac{3-2y}{5} \Rightarrow 5\mid 3-2y \Rightarrow 3-2y = 5k \Rightarrow y = \dfrac{3-5k}{2} = \dfrac{3-k}{2} - 2k \Rightarrow 2\mid 3-k \Rightarrow 3-k = 2h \Rightarrow k = 3-2h \Rightarrow 3 - 5k = 3 - 5(3-2h) = -12 + 10h \Rightarrow y = \dfrac{3-5k}{2} = \dfrac{-12+10h}{2} = 5h-6 \Rightarrow n = 13y+7 = 13(5h-6)+7 = 65h-71 = 65(h-1) - 6 = 65t-6 = 65(t-1) + 59=65m+59, m \in \mathbb{Z}$
</s>