Finding modular inverses

Trial. $7\cdot2=14\equiv3$, $7\cdot3=21\equiv10$, $7\cdot4=28\equiv6$, $7\cdot5=35\equiv2$, $7\cdot8=56\equiv1$. So $7^{-1}\equiv8\pmod{11}$.

Bézout. $\gcd(7,11)=1$: $11=1\cdot7+4$, $7=1\cdot4+3$, $4=1\cdot3+1$, so $1=4-3=4-(7-4)=2\cdot4-7=2(11-7)-7=2\cdot11-3\cdot7$. So $7\cdot(-3)\equiv1\pmod{11}$, i.e. $7^{-1}\equiv-3\equiv8\pmod{11}$.

Conclusion. $7^{-1}\equiv8\pmod{11}$.