Question

Solve the following simultaneous equations by using Cramer's method.​

Answer 1

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

Given expression is

$\rm :\longmapsto\:\dfrac{x + y - 8}{2} = \dfrac{x + 2y - 14}{3} = \dfrac{3x - y}{4}$

Taking first and second member, we have

$\rm :\longmapsto\:\dfrac{x + y - 8}{2} = \dfrac{x + 2y - 14}{3}$

$\rm :\longmapsto\:3x + 3y - 24 = 2x + 4y - 28$

$\bf\implies \:x - y = - 4 - - - (1)$

Now, Taking second and third member, we have

$\rm :\longmapsto\: \dfrac{x + 2y - 14}{3} = \dfrac{3x - y}{4}$

$\rm :\longmapsto\:4x + 8y - 56 = 9x - 3y$

$\bf\implies \:5x - 11y = - 56 - - - (2)$

Now we have equations as

$\red{\rm :\longmapsto\:x - y = - 4}$

and

$\red{\rm :\longmapsto\:5x - 11y = - 56}$

The matrix form of the above equation is

$\rm :\longmapsto\:\bigg[ \begin{matrix}1& - 1 \\ 5& - 11 \end{matrix} \bigg]\begin{gathered}\sf \left[\begin{array}{c}x\\y\end{array}\right]\end{gathered} = \begin{gathered}\sf \left[\begin{array}{c} - 4\\ - 56\end{array}\right]\end{gathered}$

where,

$\red{\rm :\longmapsto\:A = \bigg[ \begin{matrix}1& - 1 \\ 5& - 11 \end{matrix} \bigg]}$

$\red{\rm :\longmapsto\:B = \begin{gathered}\sf \left[\begin{array}{c} - 4\\ - 56\end{array}\right]\end{gathered}}$

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

So,

$\rm :\longmapsto\: |A| = \begin{array}{|cc|}\sf 1 &\sf - 1 \\ \sf 5 &\sf - 11 \\\end{array} = - 11 + 5 = - 6$

$\bf\implies \: |A| \: \ne \: 0$

It means, system of equations is consistent having unique solution.

So, Consider

$\rm :\longmapsto\: D_1 = \begin{array}{|cc|}\sf - 4 &\sf - 1 \\ \sf - 56 &\sf - 11 \\\end{array}$

$\rm \:  =  \:44 - 56$

$\rm \:  =  \: - \: 12$

Now, Consider

$\rm :\longmapsto\: D_2 = \begin{array}{|cc|}\sf 1 &\sf - 4 \\ \sf 5 &\sf - 56 \\\end{array}$

$\rm \:  =  \: - 56 - ( - 20)$

$\rm \:  =  \: - 56 + 20$

$\rm \:  =  \: - 36$

Hence,

$\bf\implies \:x = \dfrac{ - 12}{ - 6} = 2$

and

$\bf\implies \:x = \dfrac{ - 36}{ - 6} = 6$

Verification :-

Given expression is

$\rm :\longmapsto\:\dfrac{x + y - 8}{2} = \dfrac{x + 2y - 14}{3} = \dfrac{3x - y}{4}$

On substituting the values of x and y, we have

$\rm :\longmapsto\:\dfrac{2+ 6 - 8}{2} = \dfrac{2 + 2(6) - 14}{3} = \dfrac{3(2)- 6}{4}$

$\rm :\longmapsto\:\dfrac{8 - 8}{2} = \dfrac{ - 12 + 12}{3} = \dfrac{6- 6}{4}$

$\rm :\longmapsto\:0 = 0 = 0$

Hence, Verified

Answer 2

Answer:

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

Given expression is

$\rm :\longmapsto\:\dfrac{x + y - 8}{2} = \dfrac{x + 2y - 14}{3} = \dfrac{3x - y}{4}$

Taking first and second member, we have

$\rm :\longmapsto\:\dfrac{x + y - 8}{2} = \dfrac{x + 2y - 14}{3}$

$\rm :\longmapsto\:3x + 3y - 24 = 2x + 4y - 28$

$\bf\implies \:x - y = - 4 - - - (1)$

Now, Taking second and third member, we have

$\rm :\longmapsto\: \dfrac{x + 2y - 14}{3} = \dfrac{3x - y}{4}$

$\rm :\longmapsto\:4x + 8y - 56 = 9x - 3y$

$\bf\implies \:5x - 11y = - 56 - - - (2)$

Now we have equations as

$\red{\rm :\longmapsto\:x - y = - 4}$

and

$\red{\rm :\longmapsto\:5x - 11y = - 56}$

The matrix form of the above equation is

$\rm :\longmapsto\:\bigg[ \begin{matrix}1& - 1 \\ 5& - 11 \end{matrix} \bigg]\begin{gathered}\sf \left[\begin{array}{c}x\\y\end{array}\right]\end{gathered} = \begin{gathered}\sf \left[\begin{array}{c} - 4\\ - 56\end{array}\right]\end{gathered}$

where,

$\red{\rm :\longmapsto\:A = \bigg[ \begin{matrix}1& - 1 \\ 5& - 11 \end{matrix} \bigg]}$

$\red{\rm :\longmapsto\:B = \begin{gathered}\sf \left[\begin{array}{c} - 4\\ - 56\end{array}\right]\end{gathered}}$

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

So,

$\rm :\longmapsto\: |A| = \begin{array}{|cc|}\sf 1 &\sf - 1 \\ \sf 5 &\sf - 11 \\\end{array} = - 11 + 5 = - 6$

$\bf\implies \: |A| \: \ne \: 0$

It means, system of equations is consistent having unique solution.

So, Consider

$\rm :\longmapsto\: D_1 = \begin{array}{|cc|}\sf - 4 &\sf - 1 \\ \sf - 56 &\sf - 11 \\\end{array}$

$\rm \:  =  \:44 - 56$

$\rm \:  =  \: - \: 12$

Now, Consider

$\rm :\longmapsto\: D_2 = \begin{array}{|cc|}\sf 1 &\sf - 4 \\ \sf 5 &\sf - 56 \\\end{array}$

$\rm \:  =  \: - 56 - ( - 20)$

$\rm \:  =  \: - 56 + 20$

$\rm \:  =  \: - 36$

Hence,

$\bf\implies \:x = \dfrac{ - 12}{ - 6} = 2$

and

$\bf\implies \:x = \dfrac{ - 36}{ - 6} = 6$

Verification :-

Given expression is

$\rm :\longmapsto\:\dfrac{x + y - 8}{2} = \dfrac{x + 2y - 14}{3} = \dfrac{3x - y}{4}$

On substituting the values of x and y, we have

$\rm :\longmapsto\:\dfrac{2+ 6 - 8}{2} = \dfrac{2 + 2(6) - 14}{3} = \dfrac{3(2)- 6}{4}$

$\rm :\longmapsto\:\dfrac{8 - 8}{2} = \dfrac{ - 12 + 12}{3} = \dfrac{6- 6}{4}$

$\rm :\longmapsto\:0 = 0 = 0$

Hence, Verified