Answer: C) x = 3, (x = 2 is an extraneous solution)
Step-by-step explanation:
[tex]\frac{x}{x-1}[/tex] - [tex]\frac{1}{x-2}[/tex] = [tex]\frac{2x-5}{x^{2}-3x+2}[/tex]
[tex]\frac{x}{x-1}[/tex] - [tex]\frac{1}{x-2}[/tex] = [tex]\frac{2x-5}{(x-1)(x-2)}[/tex]
Restrictions: Denominator cannot equal zero (x - 1 ≠ 0) and (x - 2 ≠ 0), so x ≠ 1 and x ≠ 2
[tex](x - 1)(x - 2)\frac{x}{x-1}[/tex] - [tex](x - 1)(x - 2)\frac{1}{x-2}[/tex] = [tex](x - 1)(x - 2)\frac{2x-5}{(x-1)(x-2)}[/tex]
x(x - 2) - 1(x - 1) = 2x - 5
x² - 2x - x + 1 = 2x - 5
x² - 3x + 1 = 2x - 5
-2x + 5 -2x + 5
x² - 5x + 6 = 0
(x - 2)(x - 3) = 0
x - 2 = 0 x - 3 = 0
x = 2 x = 3
NOTE: x = 2 is an extraneous solution because it is one of the restricted values.