Question

question number B
please give answer ASAP​

Answer 1

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

Given rational expression is

$\rm :\longmapsto\:\dfrac{x}{x - 1} + \dfrac{x + 4}{x + 2} = \dfrac{29}{10}$

On taking LCM on LHS, we get

$\rm :\longmapsto\:\dfrac{x(x + 2) + (x + 4)(x - 1)}{(x - 1)(x + 2)}= \dfrac{29}{10}$

$\rm :\longmapsto\:\dfrac{ {x}^{2} + 2x + {x}^{2} + 4x - x - 4}{ {x}^{2} + 2x - x - 2}= \dfrac{29}{10}$

$\rm :\longmapsto\:\dfrac{ 2{x}^{2} + 5x - 4}{ {x}^{2} + x- 2}= \dfrac{29}{10}$

$\rm :\longmapsto\:29( {x}^{2} + x - 2) = 10( {2x}^{2} + 5x - 4)$

$\rm :\longmapsto\:29{x}^{2} + 29x - 58 = {20x}^{2} + 50x - 40$

$\rm :\longmapsto\:29{x}^{2} + 29x - 58 - {20x}^{2} - 50x + 40 = 0$

$\rm :\longmapsto\:9{x}^{2} - 21x - 18 = 0$

$\rm :\longmapsto\:3(3{x}^{2} - 7x - 6) = 0$

$\rm :\longmapsto\:3{x}^{2} - 7x - 6 = 0$

$\rm :\longmapsto\:3{x}^{2} - 9x + 2x- 6 = 0$

$\rm :\longmapsto\:3x(x - 3) + 2(x - 3) = 0$

$\rm :\longmapsto\:(x - 3)(3x + 2) = 0$

$\bf\implies \:x = 3 \: \: or \: \: x = - \: \dfrac{2}{3}$

Verification :-

Given expression is

$\rm :\longmapsto\:\dfrac{x}{x - 1} + \dfrac{x + 4}{x + 2} = \dfrac{29}{10}$

When x = 3

Consider LHS

$\rm :\longmapsto\:\dfrac{x}{x - 1} + \dfrac{x + 4}{x + 2}$

$\rm \: = \: \:\dfrac{3}{3 - 1} + \dfrac{3 + 4}{3 + 2}$

$\rm \: = \: \:\dfrac{3}{2} + \dfrac{7}{5}$

$\rm \: = \: \:\dfrac{15 + 14}{10}$

$\rm \: = \: \:\dfrac{29}{10}$

Hence, Verified

$\bf \: When \: x = - \: \dfrac{2}{3}$

Consider LHS

$\rm :\longmapsto\:\dfrac{x}{x - 1} + \dfrac{x + 4}{x + 2}$

$\rm \: = \:\dfrac{- \: \dfrac{2}{3}}{- \: \dfrac{2}{3} - 1} + \dfrac{- \: \dfrac{2}{3}+ 4}{- \: \dfrac{2}{3} + 2}$

$\rm \: = \:\dfrac{- \: \dfrac{2}{3}}{\: \dfrac{ - 2 - 3}{3}} + \dfrac{- \: \dfrac{2 + 12}{3}}{ \: \dfrac{ - 2 + 6}{3}}$

$\rm \: = \: \:\dfrac{ - 2}{ - 5} + \dfrac{10}{4}$

$\rm \: = \: \:\dfrac{2}{5} + \dfrac{5}{2}$

$\rm \: = \: \:\dfrac{4 + 25}{10}$

$\rm \: = \: \:\dfrac{29}{10}$

Hence, Verified