Question 2 On Congruence Solve the following system of congruences: $$x \equiv 2 \pmod 5$$ $$x \equiv 1 \pmod8$$ $$x \equiv 7 \pmod 9$$ $$\quad x \equiv -3 \pmod {11}$$   
i have no idea how to go about starting this, any help would be much appreciated.
 A: For example
$$x=2\pmod 5\Longleftrightarrow \,\exists\,\, k\in\Bbb Z\,\,\,s.t.\,\,x=2+5k$$
$$x=1\pmod 8\Longleftrightarrow\,\,\exists\,\,m\in\Bbb Z\,\,\,s.t.\,\,\,x=1+8m$$
and etc. 
You can continue with this, form some equations and find some solution...or else you can use the powerful and nice CRT = Chinese Remainder Theorem, and in particular its proof , and do it faster, cleaner and nicer.
A: Another option is using the linear diophantine equation $5a-8b=-1$.
Because $x=5a+2=8b+1$.
This will give a certain condition for $a$ of the shape $a=8k+k_a$, where $k_a$ is a constant. So $x=40k+(5k_a+2)$, or $x\equiv 5k_a+2\pmod{40}$.
This reduces the first two equations to only one. If you repeat this twice you will finally obtain something of the shape $x\equiv p\pmod{q}$
A: Solving the first and second equations simultaneously, you get x= 17 mod 40. Solving the third and fourth equations simultaneously, you get x = 52 mod 99.
Solving these two results simultaneously, you get x = 1537 mod 3960 which gives you all possible solutions.
A: Hint
Consider the solution to
$$
\begin{align}
x&\equiv1\pmod{5}\\
x&\equiv0\pmod{8}\\
x&\equiv0\pmod{9}\\
x&\equiv0\pmod{11}\\
\end{align}\tag{1}
$$
Suppose we can solve
$$
5a+792b=1\tag{2}
$$ then $x\equiv792b\pmod{3960}$ would satisfy $(1)$. We can use the Euclid-Wallis Algorithm to solve $(2)$:
$$
\begin{array}{r}
&&158&2&2\\\hline
1&0&1&\color{#C00000}{-2}&\color{#C00000}{5}\\
0&1&-158&317&-792\\
792&5&2&1&0\\
&&&{\uparrow}&{\star}
\end{array}\tag{3}
$$
$(3)$ tells us that $b\equiv-2\equiv3\pmod{5}$ satisfies $(2)$. Thus, $x\equiv2376\pmod{3960}$ satisfies $(1)$.
There are three other sets of equations corresponding to $(1)$ that can be solved and then combined to get the solution to your problem.
