Edit: I can't actually see how Chinese remainder theorem works here, if we had only $x$ on the left of each equation I can see how I could work it, but we don't. I can't seem to reduce it down to just $x$ either.
Note: this is a textbook problem.
Describe all integers that satisfy the four following equations. I imagine there will only be one, a I think I can attack this problem using a method I read about on Wikipedia called the chinese remainder theorem, is this a good place to start, or is there a trivial method for solving this?
$$\begin{align*} \\3{}x + 4 ≡ {}5 \pmod {7}. \\4x {}+ 5 ≡ 6 \pmod {9}. \\6x + {}1 ≡{} 7 \pmod {11}.\\7x + 2{} ≡ 8 \pmod {1{}3}. \end{align*}$$
I will edit in my attempt of the chinese remainder theorem after I get confirmation that it is optimal(if it is), thank you all!
HINT: For $3x+4\equiv5\mod7$, note that we can first bring the $4$ over to the RHS to get $$3x\equiv1\mod 7.$$ Now, we would like to "divide" the equation by $3$ to rid us of the pesky $3$ on the LHS. To do so, we need to find an inverse for $3$ modulo $7$. To do so, we would need to turn to the Euclidean algorithm with $3$ and $7$ because it would give us $u$ and $v$ such that $$3u+7v=1\implies 3u=1\mod7.$$
Think you can take it from here? (Mouse over to see what $3x\equiv1\mod 7$ becomes.)
I'll leave the CRT bit (which is the fun bit) to you. :)