Question

solve: x/x-1 + x-1/x = 2½​

Answer 1

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

$\sf \longmapsto \dfrac{x}{x - 1} + \dfrac{x - 1}{x} = \dfrac{5}{2}$

Taking LCM of x-1 and x

$\sf \longmapsto \dfrac{ {x}^{2} + {(x - 1)}^{2} }{x(x - 1)} = \dfrac{5}{2}$

Identity used here

(a-b)²=a²+b²-2ab

$\sf \longmapsto \dfrac{ {x}^{2} + {x}^{2} + 1 - 2x }{ {x}^{2} - x} = \dfrac{5}{2}$

$\sf \longmapsto \dfrac{2 {x}^{2} - 2x + 1 }{ {x}^{2} - x} = \dfrac{5}{2}$

Cross multiply to both sides

$\sf \longmapsto2(2 {x}^{2} - 2x + 1) = 5( {x}^{2} - x)$

$\sf \longmapsto4 {x}^{2} - 4x + 2 = 5 {x}^{2} - 5x$

Making like terms together:

$\sf \longmapsto5 {x}^{2} - 4 {x}^{2} - 5x + 4x - 2 = 0$

$\sf \longmapsto {x}^{2} - x - 2 = 0$

Now, factorise it

$\sf \longmapsto {x}^{2} - 2x + x - 2 = 0$

$\sf \longmapsto \: x(x - 2) + 1(x - 2) = 0$

$\sf \longmapsto(x + 1)(x - 2) = 0$

$\sf \bold{x = - 1 \:,\:x = 2}$

Check:

At x= -1

$\sf \longmapsto \dfrac{ - 1}{ - 1 - 1} ( + ) \dfrac{ - 1 - 1}{ - 1}$

$\sf \longmapsto \dfrac{1}{2} + 2$

$\sf \longmapsto \dfrac{5}{2}$

Now ,at x=2

$\sf \longmapsto \dfrac{2}{2 - 1} + \dfrac{2 - 1}{2}$

$\sf \longmapsto2 + \dfrac{1}{2}$

$\sf \longmapsto \dfrac{5}{2}$

Hence, Equation is satisfied at both values of x .