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

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

### 1 thought on “5\x+y-2\x-y=-1 <br /> ,15/x+y+7/x-y=10 <br /> by cross multiplication”

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

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

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

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

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

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

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

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