Question

Which value(s) of x are solution(s) of the equation below

Answer 1

Answer:

x =-1 , x=4

Step-by-step explanation:

$\frac{1}{x-4}+\frac{x}{x-2}=\frac{2}{x^2-6x+8}\\\\\mathrm{Multiply\:by\:LCM=}\left(x-4\right)\left(x-2\right)\\\\\frac{1}{x-4}\left(x-4\right)\left(x-2\right)+\frac{x}{x-2}\left(x-4\right)\left(x-2\right)=\frac{2}{x^2-6x+8}\left(x-4\right)\left(x-2\right)\\\\Simplify\\\\x-2+x\left(x-4\right)=2\\\\Solve\\\\x=4,\:x=-1$

Answer 2

Answer:

$\frac{1}{x - 4} + \frac{x}{x - 2} = \frac{2}{ {x}^{2} - 6x + 8} \\ \\ \frac{1}{x - 4} + \frac{x}{x - 2} = \frac{2}{(x - 4)(x - 2)} \\ \\ \frac{(x - 2) + x(x - 4)}{(x - 4)(x - 2)} = \frac{2}{(x - 4)(x - 2)} \\ \\ x - 2 + {x }^{2} - 4x = 2 \\ \\ {x}^{2} - 3x - 4 = 0 \\ (x - 4)(x + 1) = 0 \\ x = 4 \: \: and \: \: x = - 1$