The solutions of the given rational equation are negative one and three.
What is a rational number?
If the value of a numerical expression is terminating then they are the rational number then they are called the rational number and if the value of a numerical expression is non-terminating then they are called an irrational number.
The rational equation is given below.
[tex]\rm \dfrac{1}{x-4} + \dfrac{x}{x-2} = \dfrac{2}{x^2-6x+8}[/tex]
By solving the equation, then we have
[tex]\begin{aligned} \dfrac{1}{x-4} + \dfrac{x}{x-2} &= \dfrac{2}{x^2-6x+8}\\\\ \dfrac{x- 2 + x(x - 4)}{(x-4)(x - 2)} &= \dfrac{2}{x^2-6x+8}\\\\\dfrac{x^2 -4x + x - 2}{x^2 - 6x + 8} &= \dfrac{2}{x^2-6x+8}\\\\x^2 - 3x - 2 &= 2\\\\x^2 - 3x - 4 &= 0\end[/tex]
Then the solution of the equation will be
x² - 3x - 4 = 0
x² - 4x + x - 4 = 0
x(x - 4) + 1 (x - 4) = 0
(x - 4)(x + 1) = 0
x = -1, 4
More about the rational number link is given below.
https://brainly.com/question/9466779
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