Question

Solve the following question by factorisation:-$\dfrac{x + 1}{x - 1} + \dfrac{x - 2}{x + 2} = 3$
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Answer 1

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

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

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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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$\sf \color{azure}\fcolorbox{pink}{black}{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: @Jay Maharashtra 3210 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:}$

Answer 2

Answer

$\frac{x + 1}{x - 1} + \frac{x - 2}{x + 2} = 3 \\ \frac{(x + 1) \times (x + 2) + (x - 1) \times (x - 2)}{(x - 1) \times (x + 2)} = 3 \\ \frac{x(x + 2) + 1(x + 2) + x(x - 2) - 1(x - 2)}{x(x + 2) - 1(x + 2)} = 3 \\ \frac{x {}^{2} + 2x + x + 2 + x {}^{2} - 2x - x + 2 }{x {}^{2} + 2x - x - 2 } = 3 \\ \frac{2x {}^{2} + 4 }{x {}^{2} + x - 2 } = 3 \\ \frac{2 + 4}{x - 2 } = 3\\ 6 = 3(x - 2) \\ 6 = 3x - 6 \\ 6 + 6 = 3x \\ 12 = 3x \\ x = \frac{12}{3} \\ x = 4$

I HOPE ITS HELP YOU