Question

Guys pls help with this sum

Answer 1

Solution:-

$\blue{ \rm \: we \: have}$

$: \implies \rm \dfrac{1}{1 + x} + \dfrac{2 - x}{3 - x} = 1 + \dfrac{6}{35}$

$\blue{ \rm \: taking \: lcm \: on \: both \: side \: }$

$\rm: \implies \dfrac{1 \times (3 - x) + (2 - x)(1 + x)}{(1 + x)(3 - x)} = \dfrac{35 + 6}{35}$

$\rm : \implies \: \dfrac{(3 - x) + (2 + 2x - x - {x}^{2} )}{(3 - x + 3x - {x}^{2} )} = \dfrac{41}{35}$

$: \implies \rm \dfrac{3 - x + 2 + x - {x}^{2} }{3 + 2x - {x}^{2} } = \dfrac{41}{35}$

$\rm : \implies \: \dfrac{5 - {x}^{2} }{3 + 2x - {x}^{2} } = \dfrac{41}{35}$

$\blue{ \rm \: using \: cross \: multiplication}$

$: \implies \rm35(5 - {x}^{2} ) = 41(3 + 2x - {x}^{2} )$

$\rm : \implies \: 175 - 35 {x}^{2} = 123 + 82x - 41 {x}^{2}$

$\rm : \implies175 - 35 {x}^{2} - 123 - 82x + 41 {x}^{2} = 0$

$\rm : \implies 6 {x}^{2} - 82x + 52 = 0$

$: \implies \rm \: 2(3x {}^{2} - 41x + 26) = 0$

$: \implies \rm \: 3x {}^{2} - 41x + 26 = 0$

$: \implies \: \rm \: 3 {x}^{2} - 39x - 2x + 26 = 0$

$\rm : \implies \: 3x( {x}^{} - 13) - 2(x - 13) = 0$

$\rm : \implies(3x - 2)(x - 13) = 0$

$\blue{ \rm \: so}$

$\rm \: x = \dfrac{2}{3} \: \: and \: x = 13$