Question 275637
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Presuming that you meant to write (x+3)/12=6/(x-3), which would mean *[tex \Large \frac{x+3}{12}\ =\ \frac{6}{x-3}], as opposed to what you wrote which means *[tex \Large x\ +\ \frac{3}{12}\ =\ \frac{6}{x}\ -\ 3], proceed as follows:


Cross-multiply:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(x\ +\ 3\right)\left(x\ -\ 3\right)\ =\ 72]


Simplify:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ 9\ =\ 72]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ 81\ =\ 0]


Finally, use the difference of two squares factorization to factor and solve the quadratic.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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