Question

5\x+y-2\x-y=-1
,15/x+y+7/x-y=10
by cross multiplication

Answer 1

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

Given pair of linear equations are

$\rm :\longmapsto\:\dfrac{5}{x + y} - \dfrac{2}{x - y} = - 1 - - - (1)$

and

$\rm :\longmapsto\:\dfrac{15}{x + y} + \dfrac{7}{x - y} = 10 - - - (2)$

Let assume that,

$\rm :\longmapsto\:\dfrac{1}{x + y} = u - - - (3)$

and

$\rm :\longmapsto\:\dfrac{1}{x - y} = v - - - (4)$

So, equation (1) and (2) can be reduced to

$\rm :\longmapsto\:5u - 2v = - 1$

and

$\rm :\longmapsto\:15u + 7v = 10$

Now, using crossing multiplication method, we have

$\begin{gathered}\boxed{\begin{array}{c|c|c|c} \bf 2 & \bf 3 & \bf 1& \bf 2\\ \frac{\qquad}{} & \frac{\qquad}{}\frac{\qquad}{} &\frac{\qquad}{} & \frac{\qquad}{} &\\ \sf - 2 & \sf - 1 & \sf 5 & \sf - 2\\ \\ \sf 7 & \sf 10 & \sf 15 & \sf 7\\ \end{array}} \\ \end{gathered}$

$\rm :\longmapsto\:\dfrac{u}{ - 20 + 7} = \dfrac{v}{ - 15 - 50} = \dfrac{ - 1}{ 35 + 30}$

$\rm :\longmapsto\:\dfrac{u}{ - 13} = \dfrac{v}{ - 65} = \dfrac{1}{ - 65}$

$\rm :\longmapsto\:\dfrac{u}{1} = \dfrac{v}{5} = \dfrac{1}{5}$

$\bf\implies \:u = \dfrac{1}{5} \: \: and \: \: v = 1$

On Substituting the values of u and v in equation (3) and equation (4), we get

$\bf\implies \: \dfrac{1}{x + y} = \dfrac{1}{5} \: \: and \: \: \dfrac{1}{x - y} = 1$

So, we get

$\rm :\longmapsto\:x + y = 5$

and

$\rm :\longmapsto\:x - y = 1$

On cross multiplication method, we get

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

$\rm :\longmapsto\:\dfrac{x}{ 1 + 5} = \dfrac{y}{ 5 -1} = \dfrac{ - 1}{ - 1 - 1}$

$\rm :\longmapsto\:\dfrac{x}{6} = \dfrac{y}{4} = \dfrac{ - 1}{ - 2}$

$\rm :\longmapsto\:\dfrac{x}{6} = \dfrac{y}{4} = \dfrac{1}{2}$

$\rm :\longmapsto\:\dfrac{x}{3} = \dfrac{y}{2} = \dfrac{1}{1}$

$\bf\implies \:x = 3 \: \: \: and \: \: \: y = 2$

$\begin{gathered}\begin{gathered}\bf\: \bf :\longmapsto\:Hence-\begin{cases} &\sf{x = 3} \\ &\sf{y = 2} \end{cases}\end{gathered}\end{gathered}$