Question 986000
<font face="Times New Roman" size="+2">


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


The two denominators have no common factors, hence the lowest common denominator is the product of the two denominators.


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


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


Multiply both sides by the denominator in the LHS.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3x^2\ -\ 2x\ -\ 1\ +\ 2x^2\ +\ x\ -\ 1\ =\ x^2\ -\ 1]


Collect like terms and put them in the LHS


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4x^2\ -\ x\ -\ 1\ =\ 0]


Solve the quadratic.  It doesn't factor so use the Quadratic Formula.


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

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \