Answer
$\begin{align}
& x=\frac{{{D}_{x}}}{D},y=\frac{{{D}_{y}}}{D},z=\frac{{{D}_{z}}}{D},\text{ where D}\ne \text{0} \\
& \\
\end{align}$.
Work Step by Step
A linear system in three variables is given by:
$\begin{align}
& {{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z={{d}_{1}} \\
& {{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z={{d}_{2}} \\
& {{a}_{3}}x+{{b}_{3}}y+{{c}_{3}}z={{d}_{3}} \\
\end{align}$
Where ${{a}_{1}},{{a}_{2}},{{a}_{3,}}{{b}_{1}}\text{,}{{\text{b}}_{2}}\text{,}{{\text{b}}_{3}}\text{,}{{\text{c}}_{1}}\text{,}{{\text{c}}_{2}}\text{and }{{\text{c}}_{3}}\text{ are coffecients and }{{\text{d}}_{1}},{{d}_{2}}\And {{c}_{3}}\text{ are constants}\text{.}$
Then,
$\begin{align}
& x=\frac{{{D}_{x}}}{D},y=\frac{{{D}_{y}}}{D},z=\frac{{{D}_{z}}}{D},\text{ where D}\ne \text{0} \\
& \\
\end{align}$
Where,
$D=\left| \begin{matrix}
{{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\
{{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\
{{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\
\end{matrix} \right|\text{ }$
These are the coefficients of variables x, y, z
${{D}_{x}}=\left| \begin{matrix}
{{d}_{1}} & {{b}_{1}} & {{c}_{1}} \\
{{d}_{2}} & {{b}_{2}} & {{c}_{2}} \\
{{d}_{3}} & {{b}_{3}} & {{c}_{3}} \\
\end{matrix} \right|$
Replace $x$ -coefficients in $D$ with the constants on the right of the three equations.
${{D}_{y}}=\left| \begin{matrix}
{{a}_{1}} & {{d}_{1}} & {{c}_{1}} \\
{{a}_{2}} & {{d}_{2}} & {{c}_{2}} \\
{{a}_{3}} & {{d}_{3}} & {{c}_{3}} \\
\end{matrix} \right|$
Replace $y$ -coefficients in $D$ with the constants on the right of the three equations.
${{D}_{z}}=\left| \begin{matrix}
{{a}_{1}} & {{b}_{1}} & {{d}_{1}} \\
{{a}_{2}} & {{b}_{2}} & {{d}_{2}} \\
{{a}_{3}} & {{b}_{3}} & {{d}_{3}} \\
\end{matrix} \right|$
Replace $z$ -coefficients in $D$ with the constants on the right of the three equations.
