Answer: x = 3
x = 2 is restricted (aka extraneous)
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: x - 1 ≠ 0 → x ≠ 1 and x - 2 ≠ 0 → x ≠ 2 (because the denominator cannot equal zero)
[tex]\frac{x}{x - 1}(x - 1)(x - 2)[/tex] - [tex]\frac{1}{x - 2}(x - 1)(x - 2)[/tex] = [tex]\frac{2x - 5}{(x - 1)(x - 2)}(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
+2 +2 +3 +3
x = 2 x = 3 These are the solutions, but x = 2 is restricted
