Question

by cramers rule
fast broz​

Answer 1

$\large\underline\purple{\bold{Solution :- }}$

$\rm :\implies\:D \: = \: \begin{array}{|cc|}</p><p>3 & - 2 \\</p><p>2 & 1 \\</p><p>\end{array}$

$\rm :\implies\:D \: = \: 4 \: + \: 3$

$\rm :\implies\: \boxed{ \green{ \bf \: D \: = \: 7}}$

$\rm :\implies\:D_x \: = \: \begin{array}{|cc|}</p><p>3 & - 2 \\</p><p>16 & 1 \\</p><p>\end{array}$

$\rm :\implies\:D_x = 3 - ( - 32)$

$\rm :\implies\:D_x = 3 + 32$

$\rm :\implies \boxed{ \red{ \bf\:D_x = 35}}$

$\rm :\implies\:D_y \: = \: \begin{array}{|cc|}</p><p>3 & 3 \\</p><p>16 & 2 \\</p><p>\end{array}$

$\rm :\implies\:D_y = 6 - 48$

$\rm :\implies\: \boxed{ \blue{ \bf \: D_y = - 42}}$

So,

$\rm :\implies\:x \: = \: \dfrac{D_x}{D} = \dfrac{35}{7} = 5$

And

$\rm :\implies\:y = \dfrac{D_y}{D} = \dfrac{ - 42}{7} = - 6$