Question

Solve by factorisation x+1/x-1 + x-2/x+3 = 3​

Answer 1

$\sf\huge {\underline{\underline{{Solution}}}}$

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

Taking LCM of fractional values

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

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

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

$\sf\implies \dfrac{2 {x}^{2} + x + 5}{ {x}^{2} + 3x - x - 3 } = 3$

$\sf\implies \dfrac{2 {x}^{2} + x + 5 }{ {x}^{2} + 2x - 3} = 3$

Cross multiply to both sides:

$\sf\implies2 {x}^{2} + x + 5 = 3( {x}^{2} + 2x - 3)$

$\sf\implies2 {x}^{2} + x + 5 = 3 {x}^{2} + 6x - 9$

Making like terms together :

$\sf\implies3 {x}^{2} - 2 {x}^{2} + 6x - x - 9 - 5 = 0$

$\sf\implies {x}^{2} + 5x - 14 = 0$

Now,our Equation is in the form of a quadratic equation so,we can easily factorise it by middle term splitting method:

$\sf\implies {x}^{2} + 7x - 2x - 14 = 0$

$\sf\implies \: x(x + 7) - 2(x + 7) = 0$

$\sf\implies(x - 2)(x + 7) = 0$

∴ The roots are x = 2 & x= -7.