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. [tex]\large\underline{\sf{Solution-}}[/tex]

    Given pair of linear equations are

    [tex]\rm :\longmapsto\:\dfrac{5}{x + y} – \dfrac{2}{x – y} = – 1 – – – (1)[/tex]

    and

    [tex]\rm :\longmapsto\:\dfrac{15}{x + y} + \dfrac{7}{x – y} = 10 – – – (2)[/tex]

    Let assume that,

    [tex]\rm :\longmapsto\:\dfrac{1}{x + y} = u – – – (3) [/tex]

    and

    [tex]\rm :\longmapsto\:\dfrac{1}{x – y} = v – – – (4) [/tex]

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

    [tex]\rm :\longmapsto\:5u – 2v = – 1[/tex]

    and

    [tex]\rm :\longmapsto\:15u + 7v = 10[/tex]

    Now, using crossing multiplication method, we have

    [tex]\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}[/tex]

    [tex]\rm :\longmapsto\:\dfrac{u}{ – 20 + 7} = \dfrac{v}{ – 15 – 50} = \dfrac{ – 1}{ 35 + 30}[/tex]

    [tex]\rm :\longmapsto\:\dfrac{u}{ – 13} = \dfrac{v}{ – 65} = \dfrac{1}{ – 65}[/tex]

    [tex]\rm :\longmapsto\:\dfrac{u}{1} = \dfrac{v}{5} = \dfrac{1}{5}[/tex]

    [tex]\bf\implies \:u = \dfrac{1}{5} \: \: and \: \: v = 1[/tex]

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

    [tex]\bf\implies \: \dfrac{1}{x + y} = \dfrac{1}{5} \: \: and \: \: \dfrac{1}{x – y} = 1[/tex]

    So, we get

    [tex]\rm :\longmapsto\:x + y = 5[/tex]

    and

    [tex]\rm :\longmapsto\:x – y = 1[/tex]

    On cross multiplication method, we get

    [tex]\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}[/tex]

    [tex]\rm :\longmapsto\:\dfrac{x}{ 1 + 5} = \dfrac{y}{ 5 -1} = \dfrac{ – 1}{ – 1 – 1}[/tex]

    [tex]\rm :\longmapsto\:\dfrac{x}{6} = \dfrac{y}{4} = \dfrac{ – 1}{ – 2}[/tex]

    [tex]\rm :\longmapsto\:\dfrac{x}{6} = \dfrac{y}{4} = \dfrac{1}{2}[/tex]

    [tex]\rm :\longmapsto\:\dfrac{x}{3} = \dfrac{y}{2} = \dfrac{1}{1}[/tex]

    [tex]\bf\implies \:x = 3 \: \: \: and \: \: \: y = 2[/tex]

    [tex]\begin{gathered}\begin{gathered}\bf\: \bf :\longmapsto\:Hence-\begin{cases} &\sf{x = 3} \\ &\sf{y = 2} \end{cases}\end{gathered}\end{gathered}[/tex]

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