Question

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!! I CANNOT RETAKE THIS!!

Answer 1

Answer: C) x = 3, (x = 2 is an extraneous solution)

Step-by-step explanation:

$\frac{x}{x-1}$ - $\frac{1}{x-2}$ = $\frac{2x-5}{x^{2}-3x+2}$

$\frac{x}{x-1}$ - $\frac{1}{x-2}$ = $\frac{2x-5}{(x-1)(x-2)}$

Restrictions: Denominator cannot equal zero (x - 1 ≠ 0) and (x - 2 ≠ 0), so x ≠ 1 and x ≠ 2

$(x - 1)(x - 2)\frac{x}{x-1}$ - $(x - 1)(x - 2)\frac{1}{x-2}$ = $(x - 1)(x - 2)\frac{2x-5}{(x-1)(x-2)}$

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.