Question

2x+3y=13and 5x-4y=-2 by cross mult
iply

Answer 1

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

Given pair of linear equations is

$\rm :\longmapsto\:2x + 3y = 13$

and

$\rm :\longmapsto\:5x - 4y = - 2$

Now, Using Cross Multiplication method, we have

$\red{\begin{gathered}\boxed{\begin{array}{c|c|c|c} \bf 2 & \bf 3 & \bf 1& \bf 2\\ \frac{\qquad}{} & \frac{\qquad}{}\frac{\qquad}{} &\frac{\qquad}{} & \frac{\qquad}{} &\\ \sf 3 & \sf 13 & \sf 2 & \sf 3\\ \\ \sf - 4 & \sf - 2 & \sf 5 & \sf - 4\\ \end{array}} \\ \end{gathered}}$

So,

$\purple{\rm :\longmapsto\:\dfrac{x}{ - 6 - ( - 52)} = \dfrac{y}{65 - ( - 4)} = \dfrac{ - 1}{ - 8 - 15} }$

$\purple{\rm :\longmapsto\:\dfrac{x}{ - 6 + 52} = \dfrac{y}{65 + 4} = \dfrac{ - 1}{ -23} }$

$\purple{\rm :\longmapsto\:\dfrac{x}{46} = \dfrac{y}{69} = \dfrac{1}{23} }$

On taking first and third member, we get

$\purple{\rm :\longmapsto\:\dfrac{x}{46} = \dfrac{1}{23} }$

$\purple{\bf\implies \:x = 2 }$

On taking second and third member, we get

$\purple{\rm :\longmapsto\: \dfrac{y}{69} = \dfrac{1}{23} }$

$\purple{\bf\implies \:y = 3 }$

So, Solution is

$\green{\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:\begin{cases} &\bf{x = 2} \\ \\ &\bf{y = 3} \end{cases}\end{gathered}\end{gathered}}$

Verification :-

Consider Equation (1), we have

$\blue{\rm :\longmapsto\:2x + 3y = 13}$

On substituting the values of x and y, we get

$\blue{\rm :\longmapsto\:2 \times 2 + 3 \times 3 = 13}$

$\blue{\rm :\longmapsto\:4 + 9 = 13}$

$\blue{\bf\implies \:13 = 13}$

Hence, Verified

Now, Consider Equation (2), we have

$\pink{\rm :\longmapsto\:5x - 4y = - 2}$

On substituting the values of x and y, we get

$\pink{\rm :\longmapsto\:5 \times 2 - 4 \times 3 = - 2}$

$\pink{\rm :\longmapsto\:10 - 12 = - 2}$

$\pink{\bf\implies \: - 2 = - 2}$

Hence, Verified