So i understand how to do this when it is just x, but now with multiples of x I am a little confused, and there's no example in my textbook of this. I just need a push in the right direction for how to get started.
Find an integer x so that:
a. $3x\equiv2\pmod5$ and $4x\equiv5\pmod7$
b. $7x\equiv11\pmod9$ and $3x\equiv22\pmod{35}$
Because $2\cdot 3 = 6$ and $6\equiv 1\bmod 5$, we have that $$3x\equiv 2\bmod 5\iff 2\cdot(3x)\equiv 2\cdot (2)\bmod 5\iff 6x\equiv 4\bmod 5\iff x\equiv 4\bmod 5$$ See if you can find the right multiplication to do for the other three congruences :)
In general, if you're solving $ax\equiv b\bmod m$ and $\gcd(a,m)=1$, you'll want to look for some $c$ to multiply both sides by to get $x$ on its own. In other words, you're looking for a $c$ such that $ac\equiv 1\bmod m$. But this is equivalent to finding a $c$ and $d$ such that $$ac+dm=1$$ which is doable by Bezout's identity / extended Euclidean algorithm.