Path needed for solving these linear equations in Zn (my example Z105) So these are two equations :
$$49x \equiv 21 \pmod {105}$$
$$64x \equiv 21 \pmod {105}$$
I should find the multiplicative inverse of $64$ and then $49$ that gets the result of $1$
so....
In the first equation i know that the $\gcd(49, 21)$ is $7$
So reducing them, I have $7x \equiv 7 \pmod {105}$
and $105 = 15 \times 7$ 
here I get lost!!

In the second equation 
I know that $64$ and $21$ are coprime with each other, so euclidean algorithm, bezout identity, smth like 64v + 21w = 1 right.. and if I found V I basically found the x... but I get lost somewhere...
first I do the $\gcd(105, 64)$ then with bezout identity I want to find v and w such that = $1$ 
and I have $25$ for the bigger term and $-41$ for the smaller term... 
but $64 x 25 + 21(-41)$ doesn't equal to $1$
again,, lost... connection not found!! :D 
please help... maybe i chose the wrong direction, and math doesn't do for me! 
 A: The inverse of $64$ modulo $105$ can be found with the Euclidean algorithm:
\begin{align}
105&=64\cdot 1+41\\
64&=41\cdot 1+23\\
41&=23\cdot 1+18\\
23&=18\cdot 1+5\\
18&=5\cdot 3+3\\
5&=3\cdot 1+2\\
3&=2\cdot 1+1
\end{align}
Thus
\begin{align}\def\c#1{\color{red}{#1}}
1&=\c{3}-\c{2}\\
&=\c{3}-(\c{5}-\c{3})=(-1)\cdot\c{5}+2\cdot\c{3}\\
&=(-1)\cdot\c{5}+2(\c{18}-3\cdot\c{5})=(-7)\cdot\c{5}+2\cdot\c{18}\\
&=-7(\c{23}-\c{18})+2\cdot\c{18}=(-7)\cdot\c{23}+(-5)\cdot\c{18}\\
&=(-7)\cdot\c{23}+(-5)(\c{41}-\c{23})=(-2)\cdot\c{23}+(-5)\cdot\c{41}\\
&=(-2)(\c{64}-\c{41})+(-5)\cdot\c{41}=(-2)\cdot\c{64}+(-3)\cdot\c{41}\\
&=(-2)\cdot\c{64}+(-3)(\c{105}-\c{64})=\c{64}+(-3)\cdot\c{105}
\end{align}
and it turns out that the inverse is $64$.
Therefore, from the second equation, we get that
$$
x\equiv 64\cdot21\equiv1344\equiv84\pmod{105}
$$
Now
$$
49\cdot 84=4116\equiv21\pmod{105}
$$
and you're done.
A: Note that you could combine these equations and get:
$15x\equiv0 (\mod 105)$ which would reduce your initial equations to $4x\equiv21 (\mod 105)$
From here, you could use the Chinese Remainder Theorem for one idea or look at $21+105k$ and find a multiple of 4 there.
$21+3*105= 21+315 = 336 = 4 * 84$
Thus, try $x\equiv84 (\mod 105)$ as one solution.
