It is much more insightful to consider the following generalization.
Theorem $\ $ The following are equivalent for integers $\rm\:c,\, m\,$.
$(1)\rm\ \ \ gcd(c,m) = 1$
$(2)\rm\ \ \ c\:$ is invertible $\rm\,(mod\ m)$
$(3)\rm\ \ \ x\to cx+d\:$ is $\:1$-$1\:$ $\rm\,(mod\ m)$
$(4)\rm\ \ \ x\to cx+d\:$ is onto $\rm\,(mod\ m)$
Proof $\ \ (1\Rightarrow 2)\ \ $ By Bezout, $\rm\: gcd(c,m) = 1\:\Rightarrow\: cd+km = 1\:$ for $\rm\:d,k\in\Bbb Z\:$ $\rm\Rightarrow\:cd\equiv 1\pmod m$
$(2\Rightarrow 3)\ \ \ \rm cx+d \equiv cy+d\:\Rightarrow\:c(x-y)\equiv 0\:\Rightarrow\:x-y\equiv 0\:$ by multiplying by $\rm\:c^{-1}$
$(3\Rightarrow 4)\ \ $ Every $1$-$1$ function on a finite set is onto.
$(4\Rightarrow 1)\ \ \ \rm x\to cx\:$ is onto, so $\rm\:cd\equiv 1,\:$ some $\rm\,d,\:$ i.e. $\rm\: cd+km = 1,\:$ some $\rm\,k,\:$ so $\rm\:gcd(c,m)=1$