Answer: C) x = 3, (x = 2 is an extraneous solution)
<u>Step-by-step explanation:</u>
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Restrictions: <em>Denominator cannot equal zero</em> (x - 1 ≠ 0) and (x - 2 ≠ 0), so x ≠ 1 and x ≠ 2
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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.