Hey there! :)
Answer:
x = -1.
Step-by-step explanation:
[tex]\frac{1}{x-4}+ \frac{x}{x-2}= \frac{2}{x^{2}-6x+8 }[/tex]
Make each fraction have a common denominator:
[tex]\frac{1(x-2)}{x^{2}-6x+8}+ \frac{x(x-4)}{x^{2}-6x+8}= \frac{2}{x^{2}-6x+8 }[/tex]
Simplify:
[tex]\frac{x-2}{x^{2}-6x+8}+ \frac{x^{2}-4x }{x^{2}-6x+8}= \frac{2}{x^{2}-6x+8 }[/tex]
Disregard the denominator and solve the numerators:
x - 2 + x² - 4x = 2
Combine like terms:
x² - 3x - 2 = 2
x² - 3x - 4 = 0
Factor:
(x - 4)(x + 1)
***Only one of these solutions works because if x = 4, the denominator of the first fraction would be 0, which is undefined. Therefore, the only possible solution is x = -1.
