Simple system of linear congruence I'm very novice on congruences so please don't scold me too much :) So I have a system of equations like this:
$$\begin{cases}23d\equiv 1 \pmod{40}\\ 73d\equiv 1 \pmod{102} \end{cases}$$
And so from (1) I get $d=40k+7$. I plug it into (2) and I have $73(40k+7)\equiv 1 \pmod{102} \rightarrow 64k\equiv -6 \pmod{102}$. 
That means $k = 51 n+27$, so plugging back to the first, we have $d=2040n+1087$ which means that any $d\equiv 1087 \pmod{2040}$ satisfies the system of equations here which is not true since 7 doesn't satisfy it while it does satisfy both of our equations. 
How can I do it, then? Can I just see that $d=40k+7$ from the first and $d=102k+7$ from the second and just calculate $\begin{cases}d=40k+7 \\ d=102k+7 \end{cases}$ which is 7? Can I then be sure it's the only solution? 
 A: There is an error in your arithmetic:
$$73*7 = 511 = 5*102 + 1$$
Thus when you substitute in $d = 40k + 7$ you should get $64k \equiv 0 \pmod{102}$ instead of $\equiv -6$.
A: $$
\begin{align}
23d+40a&=1\tag{1}\\
73d+102b&=1\tag{2}
\end{align}
$$
$73\times(1)-23\times(2)$ yields
$$
2920a-2346b=50\tag{3}
$$
Solving $(3)$ using the Euclid-Wallis Algorithm:
$$
\begin{array}{rrrrrrr|r}
&&1&4&11&2&12\\\hline
1&0&1&-4&45&\color{#C00000}{-94}&\color{#00A000}{1173}&\color{#0000FF}{-4}\\
0&1&-1&5&-56&\color{#C00000}{117}&\color{#00A000}{-1460}&\color{#0000FF}{5}\\
2920&2346&574&50&24&\color{#C00000}{2}&\color{#00A000}{0}&\color{#0000FF}{50}\tag{4}
\end{array}
$$
Adding $2$ times the green column to $25$ times the red column gives the blue column, which tells
$$
2920(-4+1173k)+2346(5-1460k)=50\tag{5}
$$
$(5)$ gives solutions to $(3)$ of $a=-4+1173k$ and $b=-5+1460k$. Plugging back into $(1)$ or $(2)$ yields
$$
d=7-2040k\tag{6}
$$
That is,
$$
d\equiv7\pmod{2040}\tag{7}
$$
A: Hint $\ $ It is much easier to solve if you factor the moduli, e.g.
$\left.\begin{eqnarray}\rm mod\ 5\!:&&\rm\ \ \ d\equiv \dfrac{1}{23}\equiv\,  \dfrac{6}3\ \equiv\ 7\\ 
\\
\rm mod\ 8\!:&&\rm\ \ \ d\equiv \dfrac{1}{23}\equiv \dfrac{1}{-1} \equiv 7\end{eqnarray}
\right\}\! \rm \iff d\equiv 7\ \ (mod\ 40)$
$\left.\begin{eqnarray}
\rm mod\  6\!:\ &&\rm\ d\equiv \dfrac{1}{73}\equiv \dfrac{1}{1}\ \equiv\ 7\\ 
\\
\rm mod\ 17\!:&&\rm\ d\equiv \dfrac{1}{73}\equiv \dfrac{35}{5}\equiv 7\end{eqnarray}
\ \right\}\!\rm \iff d\equiv 7\ \ (mod\ 102)$
