Solve the simultaneous systems of congruences for $x \equiv 10\pmod{60}, \;x\equiv 80\pmod{350}$

Question

Solve the simultaneous systems of congruences for $x \equiv 10\pmod{60}, x\equiv 80\pmod{350}$

So the solution starts with $\gcd(60,350)=10$ then $10\mid 80-10$ because this is a requirement for there to be a solution to $x$. So then we know because so we know that $10+60k\equiv 80\pmod{350}$ which leads to $60k\equiv 70\pmod{350}$. Then somehow they jump to $k\equiv7\pmod{35}$, which I don't know how they got to that.

Anyways if I skip that part you end up with
$60k\equiv 70(mod350)$
We can use Diophantine linear eq to solve this.
$60k+350y=70$
Then solving for that I got 7+(350/10)t, $t\in Z$

so that's solving for k so then the solution is 10+60(7+35t).

Answer 1

The steps are as follows

$60k\equiv 70\pmod {350}$

$6k\equiv 7\pmod {35}$

$36k\equiv 42\pmod {35}$

$k\equiv 7\pmod {35}$

Answer 2

Alternatively: $$\begin{cases}x \equiv 10\pmod{60} \\ x\equiv 80\pmod{350}\end{cases} \Rightarrow \begin{cases}x = 60m+ 10 \\ x=350n+ 80\end{cases} \Rightarrow 60m-350n=70 \Rightarrow \\ 6m-35n=7 \Rightarrow 6m=7(5n+1)\Rightarrow 6(7m_1)=7(5n+1) \Rightarrow \\ 6m_1=5n+1 \Rightarrow m_1=1+5k,n=1+6k\\ x=350n+80=350(1+6k)+80 =2100k+430.$$