Question

if,1/(x-3)+1/(x-1)=5/(x-1)-2/(x-2),then x=?​

Answer 1

Answer:

$\frac{1}{x - 3} + \frac{1}{x - 1} = \frac{5}{x - 1} - \frac{2}{x - 2} \\ \frac{(x - 1) + (x - 3)}{(x - 3)(x - 1)} = \frac{5(x - 2) - 2(x - 1)}{(x - 1)(x - 2)} \\ \frac{x - 1 + x - 3}{x(x - 1) - 3(x - 1)} = \frac{5x - 10 - 2x + 2}{x(x - 2) - 1(x - 2)} \\ \frac{ x + x - 1 - 3}{ {x}^{2} - x - 3x + 3} = \frac{5x - 2x - 10 + 2}{ {x}^{2} - 2x - x + 2} \\ \frac{2x - 4}{ {x}^{2} - 4x + 3 } = \frac{3x - 8}{ {x}^{2} - 3x + 2} \\ (3x - 8)( {x}^{2} - 4x + 3) = (2x - 4)( {x}^{2} - 3x + 2) \\ 3x( {x}^{2} - 4x + 3) - 8( {x}^{2} - 4x + 3) = 2x( {x}^{2} - 3x + 2) - 4( {x}^{2} - 3x + 2) \\ 3 {x}^{3} - 12 {x}^{2} + 9x - 8 {x}^{2} + 32x - 24 = 2 {x}^{3} - 6 {x}^{2} + 4x - 4 {x}^{2} + 12x - 8 \\ 3 {x}^{3} - 2 {x}^{3} - 12 {x}^{2} - 8 {x}^{2} + 6 {x}^{2} + 4 {x}^{2} + 9x + 32x - 4x - 12x - 24 + 8 = 0 \\ {x}^{3} - 10 {x}^{2} + 25x - 16 = 0$