Solution: Assume there be a system of n simultaneous linear equations and with ‘n’ unknown given by

${{a}_{11}}{{X}_{1}}+{{a}_{12}}{{X}_{2}}+…+{{a}_{1n}}{{X}_{n}}={{b}_{1}}$

${{a}_{21}}{{X}_{1}}+{{a}_{22}}{{X}_{2}}+…+{{a}_{2n}}{{X}_{n}}={{b}_{2}}$

${{a}_{n1}}{{X}_{1}}+{{a}_{n2}}{{X}_{2}}+…+{{a}_{nn}}{{X}_{n}}={{b}_{n}}$

$LetD=\left| \begin{matrix}

{{a}_{11}} & {{a}_{12}} & \cdots  & {{a}_{1n}}  \\

{{a}_{21}} & {{a}_{22}} & \cdots  & {{a}_{2n}}  \\

\vdots  & \vdots  & {} & \vdots   \\

{{a}_{n1}} & {{a}_{n1}} & \cdots  & {{a}_{nn}}  \\

\end{matrix} \right|$Let ${{D}_{1}}$ be the determinant obtained from D after replacing the ${{j}^{th}}$column by

$\left| \begin{matrix}

{{b}_{1}}  \\

{{b}_{2}}  \\

\vdots   \\

{{b}_{n}}  \\

\end{matrix} \right|$Then,

${{X}_{1}}=\frac{{{D}_{1}}}{D},{{X}_{2}}=\frac{{{D}_{2}}}{D},…{{X}_{n}}=\frac{{{D}_{n}}}{D}$provided that $D\ne 0$

$9x+5y=10$

$3y-2x=8$

So by comparing with the theorem, let’s find $D,{{D}_{1}}And{{D}_{2}}$

$\Rightarrow D=\left| \begin{matrix}

9 & 5  \\

-2 & 3  \\

\end{matrix} \right|$

Solving for the determinant, expanding along 1^st row

$\Rightarrow D=3(9)-(5)(-2)$

$\Rightarrow D=27+10$

$\Rightarrow D=37$

Again,

$\Rightarrow {{D}_{1}}=\left| \begin{matrix}

10 & 5  \\

8 & 3  \\

\end{matrix} \right|$

Solving for the determinant, expanding along 1^st row

$\Rightarrow {{D}_{1}}=10(3)-(8)(5)$

$\Rightarrow {{D}_{1}}=30-40$

$\Rightarrow {{D}_{1}}=-10$

$\Rightarrow {{D}_{2}}=\left| \begin{matrix}

9 & 10  \\

-2 & 8  \\

\end{matrix} \right|$

Solving for the determinant, expanding along 1^st row

⇒ ${{D}_{2}}\text{ }\!\!~\!\!\text{ }=9(8)-(10)(-2)$

⇒ ${{D}_{2}}=72+20$

⇒ ${{D}_{2}}=92$

Therefore, by Cramer’s Rule, we will get

$\Rightarrow X=\frac{{{D}_{1}}}{D}$

$\Rightarrow X=\frac{-10}{37}$

And

$\Rightarrow Y=\frac{{{D}_{2}}}{D}$

$\Rightarrow Y=\frac{92}{37}$