Question 500728
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{x\ -\ 5}{x\ +\ 4}\ =\ \frac{x\ -\ 4}{x\ +\ 6}] 


Multiply both sides by 1 - on the left use *[tex \Large \frac{x\ +\ 6}{x\ +\ 6}] and on the right use *[tex \Large \frac{x\ +\ 4}{x\ +\ 4}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\frac{x\ -\ 5}{x\ +\ 4}\right)\left(\frac{x\ +\ 6}{x\ +\ 6}\right)\ =\ \left(\frac{x\ -\ 4}{x\ +\ 6}\right)\left(\frac{x\ +\ 4}{x\ +\ 4}\right)]


Next, multiply both sides by *[tex \Large (x\ +\ 6)(x\ +\ 4)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ -\ 5)(x\ +\ 6)\ =\ (x\ -\ 4)(x\ +\ 4)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ x\ -\ 30\ =\ x^2\ -\ 16]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ 14]


And since this potential solution does NOT cause any of the original denominators to be zero, this is the solution set.


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