Question

Solve the linear equation by Cramer’s Rule method. 6x – 3y = -10 ; 3x + 5y -8 =0​

Answer 1

$\large\underline{\sf{Solution-}}$

Given pair of linear equations are

6x - 3y = - 10

and

3x + 5y - 8 = 0

can be rewritten as

3x + 5y = 8.

So, given pair of linear equation in matrix form can be represented as

$\rm :\longmapsto\:\bigg[ \begin{matrix}6& - 3 \\ 3&5 \end{matrix} \bigg]\begin{gathered}\sf \left[\begin{array}{c}x\\y\end{array}\right]\end{gathered} = \begin{gathered}\sf \left[\begin{array}{c} - 10\\8\end{array}\right]\end{gathered}$

where,

$\rm :\longmapsto\:A = \bigg[ \begin{matrix}6& - 3 \\ 3&5 \end{matrix} \bigg]$

$\rm :\longmapsto\:X = \begin{gathered}\sf \left[\begin{array}{c}x\\y\end{array}\right]\end{gathered}$

$\rm :\longmapsto\:B = \begin{gathered}\sf \left[\begin{array}{c} - 10\\8\end{array}\right]\end{gathered}$

Now, Consider

$\rm :\longmapsto\: |A| = \begin{array}{|cc|}\sf 6 &\sf - 3 \\ \sf 3 &\sf 5 \\\end{array}$

$\rm \:  =  \:30 - ( - 9)$

$\rm \:  =  \:30 + 9$

$\rm \:  =  \:39$

$\bf\implies \: |A| = 39$

It means system of equations is consistent having unique solution.

Now, Consider

$\rm :\longmapsto\: D_1 = \begin{array}{|cc|}\sf - 10 &\sf - 3 \\ \sf 8 &\sf 5 \\\end{array}$

$\rm \:  =  \: - 50 - ( - 24)$

$\rm \:  =  \: - 50 + 24$

$\rm \:  =  \: - 26$

$\bf\implies \:D_1 = - 26$

Now, Consider

$\rm :\longmapsto\: D_2 = \begin{array}{|cc|}\sf 6 &\sf - 10 \\ \sf 3 &\sf 8 \\\end{array}$

$\rm \:  =  \:48 - ( - 30)$

$\rm \:  =  \:48 + 30$

$\rm \:  =  \:78$

$\bf\implies \:D_2 = 78$

Hence,

$\rm :\longmapsto\:x = \dfrac{D_1}{ |A| } = \dfrac{ - 26}{39} = - \dfrac{2}{3}$

and

$\rm :\longmapsto\:y = \dfrac{D_2}{ |A| } = \dfrac{78}{39} = 2$