How does Cramer's rule work? I know Cramer's rule works for 3 linear equations. I know all steps to get solutions. But I don't know why (how) Cramer's rule gives us solutions? 

Why do we get $x=\frac{\Delta_1}\Delta$ and $y$ and $z$ in the same way?

I want to know how these steps give us solutions?
It's actually simple; I explain it here in two variables, but the principle is the same.

Say you have an equation
$$\begin{pmatrix}a&amp;b\\c&amp;d\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}p\\q \end{pmatrix}$$

Now you can see that the following holds  

$$\begin{pmatrix}a&amp;b\\c&amp;d\end{pmatrix}\begin{pmatrix}x&amp;0\\y&amp;1\end{pmatrix}=\begin{pmatrix}p&amp;b\\q &amp;d\end{pmatrix}$$

Finally just take the determinant of this last equation; $\det$ is multiplicative so you get $$\Delta x=\Delta_1$$
Given a linear system $\rm A x = b$, where $\mathrm A \in \mathbb R^{n \times n}$ is invertible and $\mathrm b \in \mathbb R^n$, let $\rm \bar{x} := A^{-1} b$ denote the (unique) solution. Replacing the $k$-th column of matrix $\rm A$ (denoted by $\mathrm a_k$) with $\rm b$ and computing the determinant of this new matrix using the matrix determinant lemma,

$$\begin{aligned} \det \left( \mathrm A + (\mathrm b - \mathrm a_k) \mathrm e_k^\top \right) &amp;= \det (\mathrm A) \cdot \left( 1 + \mathrm e_k^\top \mathrm A^{-1} (\mathrm b - \mathrm a_k) \right)\\ &amp;= \det (\mathrm A) \cdot \left( 1 + \mathrm e_k^\top \mathrm A^{-1} \mathrm b - \mathrm e_k^\top \mathrm A^{-1} \mathrm a_k \right)\\ &amp;= \det (\mathrm A) \cdot \left( 1 + \mathrm e_k^\top \mathrm{\bar{x}} - \mathrm e_k^\top \mathrm e_k \right)\\ &amp;= \det (\mathrm A) \cdot \bar{x}_k\end{aligned}$$

and, thus, we obtain Cramer's rule

$$\bar{x}_k = \frac{\det \left( \mathrm A + (\mathrm b - \mathrm a_k) \mathrm e_k^\top \right)}{\det (\mathrm A)}$$