Solving simultaneous congruences with the Chinese remainder theorem Solve the following system of congruences using the Chinese remainder theorem:
$$\begin{align*}
2x &\equiv 3 \pmod{7} \\
x &\equiv 4 \pmod{6} \\
5x &\equiv 50 \pmod{55}
\end{align*}
$$
I was a little confused how to reduce the congruences into a form where the Chinese remainder theorem is applicable.
 A: *

*The first congruence takes the form
$$2x \equiv 3 \pmod 7$$
so we want to find the multiplicative inverse of $2$, modulo $7$.
You can use Euclid's algorithm for computing GCDs, or just think about it, and see that $4$ is the multiplicative inverse of $2$, as
$$2 \times 4 \equiv 1 \pmod{7}$$
So now we multiply the congruence through by $4$, and we get
$$x \equiv 3 \times 4 \equiv 5 \pmod 7$$
which is the form we want for the CRT.

*This congruence is already in the form we want.

*The third congruence is
$$5x \equiv 50 \pmod {55}$$
Since the coefficient of $x$, the remainder and the modulus all have common factor $5$, we can divide through by this to get the congruence
$$x \equiv 10 \pmod {11}$$
So now this congruence is in a suitable form to apply CRT.
(For proof that this works, observe that
$$55 \mid (5x - 50)
\iff 5 \times 11 \mid 5\,(x - 10)
\iff 11 \mid x - 10.)
$$
Now you have the three congruences
$$
\begin{align*}
x &\equiv 5 \pmod 7 \\
x &\equiv 4 \pmod 6 \\
x &\equiv 10 \pmod {11}
\end{align*}
$$
the standard form, which I assume you already know how to solve.
A: The congruence $2x\equiv 3\pmod{7}$ holds if and only if $x\equiv 5\pmod{7}$. And $5x\equiv 50\pmod{55}$ if and only if $x\equiv 10\pmod{11}$. Now the problem is in "standard" form. 
Remark: To get the first result, we can multiply both sides of $2x\equiv 3\pmod{7}$ by the inverse of $2$ modulo $7$. Note that $4$ is the inverse to $2$. Or more simply we replace the $3$ by $10$, and divide by $2$.
The second result comes from the general fact that if $k\ne 0$, then $ka\equiv kb\pmod{km}$ if and only if $a\equiv b\pmod{m}$. 
A: Solving x = 5(mod 7) and x = 4(mod 6) simultaneously, you get x = 40(mod 42).
Solving this result and x = 10(mod 11) simultaneously, you get x = 208(mod 462)
Which is the final answer.
A: $$2x≡3(mod 7)=> 2x≡3+7(mod 7)$$
$$x≡4(mod 6)$$
$$5x≡50(mod 55) => x≡10(mod 11)$$
3 linear congruence formed now apply CRT
$$m=7*6*11=462$$
$$M_1=66,M_2=77,M_3=42$$
$$66x≡1(mod 7) ; b_1=5$$
$$77x≡1(mod 6) ; b_2=-1$$
    $$42x≡1(mod 11) ; b_3=5$$
$$x≡5*5*66+77*(-1)*(4)+42*5*10(mod 462)$$
$$x≡3442 (mod 462)$$
$$x≡208(mod 462)$$
$$x=(462)t+208;   t={0,1,2,3...}$$
