Solve simultaneous systems of congruences $x\equiv 10 \pmod{60}$ and $x\equiv 80 \pmod{350}$ Solve simultaneous systems of congruences $x\equiv 10 \pmod{60}$ and $x\equiv 80 \pmod{350}$
How does one solve this using CRT? Because it has duplicate primes in its factorization? I got
$$350=5*5*2*7$$
$$60=2*2*3*5$$
So I tried CRT with
$x\equiv 1\pmod3, x\equiv0\pmod5, x\equiv0 \pmod2, x\equiv3 \pmod7$ and got the wrong answer.
or should it be
$x\equiv 1\pmod3, x\equiv5\pmod{25}, x\equiv2\pmod4, x\equiv3\pmod7$
 A: Your work will be greatly simplify using the formula for the solutions of a system of two simultaneous congruences with coprime moduli $a$ and $b$ based on Bézout's identity
$$ua+vb=1,\qquad(u,v\in\mathbf Z):$$
$$\begin{cases}
x\equiv \alpha\mod a\\x\equiv\beta\mod b
\end{cases}\iff x\equiv\beta\,ua+\alpha\,vb\mod ab.$$
Now your congruences are equivalent to $\begin{cases}x'\equiv 1\mod 6\\x'\equiv8\mod 35 \end{cases}$ and $x\equiv 10x'\mod60\cdot 350$.
Added: 
The given congruences mean there exist $k,l \in\mathbf Z$ such that  $x=10+60k=80+350l$, i.e. $x=10(1+6k)=10(8+35l)$, so $1+6k=8+35l$, which I denote $x'$, whence the simplified system of congruences and the relation between $x$ and $x'$. congruences
A: When the moduli are not coprime there may not be a solution.  With a small system, it's probably better to solve in the following fashion so that you can see when and why a contradiction arises:
From the first congruence you have $x=10+60y$.  Plug that into the second congruence to get 
$$10+60y \equiv 80 \pmod{350}$$ 
$$60y \equiv 70 \pmod{350}$$
$$6y \equiv 7 \pmod{35}$$
$6$ is its own inverse mod $35,$ so we have
$$y \equiv 42 \equiv 7 \pmod{35}$$ or $y=42+35k$.
Then $x = 10 + 60 (42 +35k) = 2530+2100 k$, or if you shift $k$ by one, $x = 430 +2100k$.
A: It should be 
$\color{magenta}{x\equiv 1\pmod3}, \color{blue}{x\equiv5\pmod{25}}, \color{magenta}{x\equiv2\pmod4,} \color{brown}{x\equiv3\pmod7}$.
From $\color{magenta}{x\equiv-2\pmod{3,4}} $ we get $x\equiv-2\equiv10\mod12.$
From $x\equiv10\pmod{12}$ and $\color{blue}{x\equiv5\pmod{25}},$ we get $x\equiv130\pmod{300}$.
From $x\equiv130\pmod{300}$ and $\color{brown}{x\equiv3\pmod7}$, we get $x\equiv430\pmod{2100}$.
