Question 332878
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You need to apply the LCD.  There is only one denominator that you can see, but there are actually two denominators in this problem.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{x}{x\ -\ 2}\ +\ \frac{3}{1}\ =\ \frac{2}{x\ -\ 2}]


Since *[tex \Large x\ -\ 2] and *[tex \Large 1] have no factors in common, the LCD is *[tex \Large (x\ -\ 2)\ \cdot\ 1\ =\ x\ -\ 2]


Apply the LCD:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{x}{x\ -\ 2}\ +\ \left(\frac{3}{1}\right)\left(\frac{x\ -\ 2}{x\ -\ 2}\right) =\ \frac{2}{x\ -\ 2}]


Simplify:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{x\ +\ 3x\ -\ 6}{x\ -\ 2}\ =\ \frac{2}{x\ -\ 2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{4x\ -\ 6}{x\ -\ 2}\ =\ \frac{2}{x\ -\ 2}]


Two fractions with identical denominators must have equal numerators.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ 4x\ -\ 6\ =\ 2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ 4x\ =\ 8]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ x\ =\ 2]


Oh oh!  Warning!! Danger, Will Robinson!  You have a single potential solution that is not in the domain of the original rational function.  If *[tex \Large x\ =\ 2], then you have a zero denominator.  Therefore the solution set is the empty set.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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