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SOVA2 [1]
3 years ago
10

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

Mathematics
1 answer:
liberstina [14]3 years ago
4 0

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

<u>Step-by-step explanation:</u>

\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: <em>Denominator cannot equal zero</em> (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

<u>     -2x + 5</u>  <u>-2x + 5 </u>

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.

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