Since $\det(A) \neq 0$ by properties of the determinant we know that $A$ is invertible.

We claim that this implies that the equation $Ax=b$ has a unique solution. Note that $A^{-1}b$ is a solution since $A(A^{-1}b)=(AA^{-1})b=b$ so we know that a solution exists.

Let $s$ be an arbitrary solution to the equation, so $As=b$ But then $s=(A^{-1}A)s=A^{-1}(As)=A^{-1}b$ so we see that $A^{-1}b$ is the only solution.

For each integer $i$ $1 \leq i \leq n$ let $a_i$ denote the $i$ column of $A$ let $e_i$ denote the $i$ column of the identity matrix $I_n$ and let $X_i$ denote the matrix obtained from $I_n$ by replacing column $i$ with the column vector $x$

We know that for any matrices $A,B$ that the $k$ column of the product $AB$ is simply the product of $A$ and the $k$ column of $B$ Also observe that $Ae_k=a_k$ for $k=1,\ldots,n$ Thus, by multiplication, we have: $$ \begin{array}{lll} AX_i & = & A (e_1,\ldots,e_{i-1},x,e_{i+1},\ldots,e_n) \\ & = & (Ae_1,\ldots,Ae_{i-1},Ax,Ae_{i+1},\ldots,Ae_n) \\ & = & (a_1,\ldots,a_{i-1},b,a_{i+1},\ldots,a_n) \\ & = & M_i \end{array} $$

Since $X_i$ is $I_n$ with column $i$ replaced with $x$ computing the determinant of $X_i$ with cofactor expansion gives: $$ \det(X_i) = (-1)^{(i+i)} x_i \det(I_{n-1}) = 1 \cdot x_i \cdot 1 = x_i$$ Thus by the multiplicative property of the determinant, $$ \det(M_i) = \det(AX_i) = \det(A) \det(X_i) = \det(A) x_i $$ and so $x_i = \frac{\det(M_i)}{\det(A)}$ as required.