Question

2x+3y=2 ; x-y/2=1/2 solve the following simultaneous equations using cramer's rule​

Answer 1

Answer:

tIn attachment ur answers is given see it

Answer 2

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

Given pair of linear equations are

$\rm :\longmapsto\:2x + 3y = 2 - - - (1)$

and

$\rm :\longmapsto\:x - \dfrac{y}{2} = \dfrac{1}{2}$

can be rewritten as

$\rm :\longmapsto\:2x - y = 1 - - - (2)$

Now, matrix form of above equations is

$\rm :\longmapsto\:A = \begin{bmatrix} 2 & 3\\ 2 & - 1\end{bmatrix}$

$\begin{gathered}\sf\rm :\longmapsto\: B=\left[\begin{array}{c}2\\1\ \\ \end{array}\right]\end{gathered}$

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

So,

$\rm :\implies\:AX = B$

Now,

$\rm :\longmapsto\: |A| = \begin{array}{|cc|}\sf 2 &\sf 3 \\ \sf 2 &\sf - 1 \\\end{array}$

$\rm \: \: = \: - 2 - 6$

$\rm \: \: = \: - 8$

$\bf\implies \: |A| = - 8 \: \ne \: 0$

System of equations is consistent having unique solution.

Consider,

$\rm :\longmapsto\:D_1 = \begin{array}{|cc|}\sf 2 &\sf 3 \\ \sf 1 &\sf - 1 \\\end{array}$

$\rm \: \: = \: - 2 - 3$

$\rm \: \: = \: - 5$

$\bf\implies \:D_1 = - \: 5$

Consider,

$\rm :\longmapsto\:D_2 = \begin{array}{|cc|}\sf 2 &\sf 2 \\ \sf 2 &\sf 1 \\\end{array}$

$\rm \: \: = \: 2 - 4$

$\rm \: \: = \: - 2$

$\bf\implies \:D_2 = - \: 2$

So,

By Cramer's Rule,

$\bf :\longmapsto\:x = \dfrac{D_1}{ |A| } = \dfrac{ - 5}{ - 8} = \dfrac{5}{8}$

$\bf :\longmapsto\:y = \dfrac{D_2}{ |A| } = \dfrac{ - 2}{ - 8} = \dfrac{1}{4}$