Question

Solve $\frac{1}{x-2} +\frac{2}{x-1} =\frac{6}{x} .$

Answer 1

SOLUTION

Given,

$\frac{1}{x - 2} + \frac{2}{x - 1} = \frac{6}{x} \\ \\ = > \frac{(x - 1) + 2(x - 2)}{(x - 2)(x - 1)} = \frac{6}{x} \\ \\ = > \frac{3x - 5}{ {x}^{2} - 3x + 2 } = \frac{6}{x} \\ \\ = > {3x}^{2} - 5x = {6x}^{2} - 18x + 12 \\ \\ = > {3x}^{2} - 13x + 12 = 0 \\ \\ = > 3 {x}^{2} - 9x - 4x + 12 = 0 \\ \\ = > 3x(x - 3) - 4(x - 3) = 0 \\ \\ = > (x - 3)(3x - 4) = 0 \\ \\ = > x = 3 \: \: or \: \: x = \frac{4}{3}$

Which are the required roots of the given equation.

Hope it helps ☺️

Answer 2

SOLUTION:-

We have,

$\frac{1}{x - 2} + \frac{2}{x - 1} = \frac{6}{x}$

So,

Explanation:

$= > \frac{1}{x - 2} + \frac{2}{x - 1} = \frac{6}{x} \\ \\ = > \frac{1(x - 1) + 2(x - 2)}{(x - 2)(x - 1)} = \frac{6}{x} \\ \\ = > \frac{x - 1+ 2x - 4}{ {x}^{2} - x - 2x + 2} = \frac{6}{x} \\ \\ = > \frac{3x - 5}{ {x}^{2} - 3x + 2} = \frac{6}{x} \\ (cross \: multiplication) \\ \\ = > x(3x - 5) = 6( {x}^{2} - 3x + 2) \\ \\ = > 3 {x}^{2} - 5x = 6 {x}^{2} - 18x + 12 \\ \\ = > 6 {x}^{2} - 3 {x}^{2} - 18 {x}^{} + 5x + 12 = 0 \\ \\ = > 3x {}^{2} - 13x + 12 = 0 \\ \\ = > 3 {x}^{2} - 9x - 4x + 12 = 0 \\ \\ = > 3x(x - 3) - 4(x - 3) = 0 \\ \\ = > ( x - 3)(3x - 4) = 0 \\ \\ = > x - 3 = 0 \: \: or \: \: 3x - 4 = 0 \\ \\ = > x = 3 \: \: or \: \: 3x = 4 \\ \\ = > x = 3 \: \: or \: \: x = \frac{4}{3 }$

Thank you.