<s>[INST] Is it possible to calculate $3^{-1}\equiv ?\pmod{10}$? If I wanted to calculate  $3^{-1}\equiv ?\pmod{10}$ would I first calcuate $3^1$ which is just 
$3\equiv 3\pmod{10}$ and then divide both sides by $3^2$ which would get $3^{-1}\equiv 3^{-1} mod{10}$ Then im not sure what to do next. My book states that $3^{-1}\equiv 7\pmod{10}$ which I don't know how they get?
 [/INST] A number $x^{-1}$ is one such that $x \cdot x^{-1} = 1$ (here $\mod 10$). In this case it is easy, because there is only 10 possibilities: $0,1,2,3,4,5,6,7,8,9$, with product being $0,3,6,9,2,5,8,1,4,7$ respectively. That means there exists, and there exists only one number such that $$x \cdot x^{-1} = 1 \mod 10.$$ There are other solutions like $17, 27, 37$ but those are $\geq 10$ or $< 0$ (curiously $-3$ is such a number, that is $3\cdot -3 = -9 = 1 \mod 10$).
The general approach can be done via the Euclidean algorithm, which for any numbers $a,b$ lets you find $x,y$ such that 
$$a x + b y = \gcd(a,b).$$
Happily $3$ and $10$ are coprime, that is $\gcd(3,10) = 1$, which allows as to run the algorithm and get $3 \cdot (-3) + 10 \cdot (1) = 1$, giving us $x = -3$, now it is enough to move $x$ to appropriate interval, that is $x+10k$ such that $0 \leq x+10k < 10$, and that gives us $7$.
I hope this helps ;-)
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