Question

Solve the following question by factorisation:-

$\dfrac{x + 1}{x - 1} + \dfrac{x - 2}{x + 2} = 3$
​

Answer 1

Given Equation

$\tt \to \: \dfrac{(x + 1)}{(x - 1)} + \dfrac{(x - 2)}{(x + 2)} = 3$

Now Take LCM

$\tt\to \: \dfrac{(x + 1)(x + 2) + (x - 2)(x - 1)}{(x - 1)(x + 2)} = 3$

$\tt \to \: \dfrac{ {x}^{2} + 2x + x + 2 + {x}^{2} - x - 2x + 2 }{ {x}^{2} + 2x - x - 2} = 3$

$\tt \to \: \dfrac{ {x}^{2} + 3x + 2 + {x}^{2} - 3x + 2}{ {x}^{2} + x - 2 } = 3$

$\tt \to \: \dfrac{ 2{x}^{2} + 4}{ {x}^{2} + x - 2} = 3$

Using Cross multiplication

$\tt \to \: 2{x}^{2} + 4 = 3( {x}^{2} + x - 2)$

$\tt \to \: 2{x}^{2} + 4 = 3 {x}^{2} + 3x - 6$

$\tt \to \: 3 {x}^{2} - 2{x}^{2} + 3x - 6 - 4 = 0$

$\tt \to \: {x}^{2} + 3x - 10 = 0$

Now Using Middle term Splitting

$\tt \to \: {x}^{2} + 5x - 2x - 10 = 0$

$\tt \to \: x(x + 5) - 2(x + 5) = 0$

$\tt \to \: (x - 2)(x + 5) = 0$

$\tt \to \: (x - 2) = 0 \: and \: (x + 5) = 0$

$\tt \to \: x = 2 \: and \: x = - 5$

Answer

$\tt \to \: x = 2 \: and \: x = - 5$

Answer 2

$\large {\pmb{\sf{Question:}}}$

$\sf\dfrac{x + 1}{x - 1} + \dfrac{x - 2}{x + 2} = 3$

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$\large{\pmb{\sf{Final \: Solution:}}}$

$\red \bigstar\color{green} \boxed{ \sf \purple{x = - 5}} \\ \red \bigstar\color{green} \boxed{ \sf \purple{x = 2}}$

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$\large \pmb{\sf{Step-by-step \; Solution:}}$

$\sf : \leadsto \: \dfrac{x + 1}{x - 1} + \dfrac{x - 2}{x + 2} = 3$

$\color{maroon}\footnotesize \boxed{ \textsf \color{green}{Taking the LCM of (x-1) and (x+2) i.e. combining the terms}}$

$\sf{: \leadsto \: \frac{(x + 1 )(x + 2) + (x - 1)(x - 2)}{(x - 1)(x + 2)} = 3 }$

$\sf{: \leadsto \: \frac{ {x}^{2} + 2 x + x + 2 + {x}^{2} - x - 2x + 2 }{ {x}^{2} + 2x - x - 2 } = 3 }$

$\sf{: \leadsto \: \frac{2 {x}^{2} + 4}{ {x}^{2} + x - 2 } = 3}$

$\color{maroon}\footnotesize \boxed{ \sf \color{green}{Transposing \: x^2 + x - 2 \: to \: another \: side}}$

$\sf : \leadsto2 {x}^{2} + 4 = 3( {x}^{2} + x - 2)$

$\sf : \leadsto 2 {x}^{2} + 4 = 3 {x}^{2} + 3x - 6$

$\color{maroon}\footnotesize \boxed{ \sf \color{green}{Transposing \:2x^2 + 4 \: to \: another \: side}}$

$\sf : \leadsto 0 = 3 {x}^{2} + 3x - 6 - 2 {x}^{2} - 4$

$\sf: \leadsto {x}^{2} + 3x - 10 = 0$

$\color{maroon}\footnotesize \boxed{ \sf \color{green}{Splitting \: the \: Middle \: Term}}$

$\sf: \leadsto {x}^{2} + 5x - 2x - 10 = 0$

$\color{maroon}\footnotesize \boxed{ \textsf \color{green}{Taking x as common from x² + 5x and 2 from 2x - 10}}$

$\sf: \leadsto x( x + 5) - 2(x + 5) = 0$

$\pink \bigstar \: \color{purple}\boxed{ \sf: \color{blue}\leadsto (x - 2 )(x + 5) = 0}$

$\pink \bigstar \: \color{purple}\boxed{\sf \color{blue}: \leadsto x = 2 \: \: \: ; \: \: \: x = -5}$

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$\sf \color{azure}\fcolorbox{pink}{black}{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: @Saanvigrover2007 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:}$