Let $a,b,m,n \in \mathbb{N}$ with $\gcd(m,n)=1$ prove $\{ x = a \pmod m,x=b \pmod n\}$ has a solution Let $a,b,m,n \in \mathbb{N}$ with $\gcd(m,n)=1$ prove $\{ x = a \pmod m,x=b \pmod n\}$ has a solution  and it is unique modulo $mn$

Definitions
$x=a \pmod m \iff m|(x-a) \iff x-a=m q_1$ where $q_1 \in \Bbb N$ 
$x=b \pmod n \iff n|(x-b) \iff x-b=n q_2$ where $q_2 \in \Bbb N$
Also, $\gcd(m,n)=1$ so $\exists  u,v \in \Bbb N$ s.t. $mu+vn=1$

Stuck on Putting it together
$x=a+m q_1$ and $x=b+nq_2$, 
Now $$\begin{aligned} x&=a+m q_1*1=a+m q_1*(um+vn)=a+mmuq_1+vn*mq_1  \\
                      x&=b+nq_2*(1)=b+nq_2(mu+vn)=b+nq_2mu+n^2 v q_2
\end{aligned}$$ 
Kind of lost at this point not sure if I took a wrong turn somewhere
$\vdots$
___=____ $\pmod {mn}$


You already have $mu+vn=1$, so $mub+vnb=b$ so $$mub\equiv b\pmod{n}$$
Similarly, you hvae $$vna\equiv a\pmod{m}$$
Now just let $$x=mub+vna$$
