Question 273701
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Please use parentheses to group things so that we don't have to guess where your numerators and denominators end and the next term starts.  I presume that you meant:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 6\ +\ \frac{4}{x\,-\,1}\ =\ \frac{5}{x\,-\,3}]


Multiply both sides by the common denominator which is the product of the two denominators.  Leave the common denominator in factored form until there are no more rational expressions:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 6(x\,-\,1)(x\,-\,3)\ +\ \frac{4(x\,-\,1)(x\,-\,3)}{x\,-\,1}\ =\ \frac{5(x\,-\,1)(x\,-\,3)}{x\,-\,3}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 6(x\,-\,1)(x\,-\,3)\ +\ 4(x\,-\,3)\ =\ 5(x\,-\,1)]


Now apply FOIL and the distributive property:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 6(x^2\ -\ 4x\ +\ 3)\ +\ 4x\ -\ 12\ =\ 5x\ -\ 5]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 6x^2\ -\ 24x\ +\ 18\ +\ 4x\ -\ 12\ =\ 5x\ -\ 5]


Collect like terms on the left:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 6x^2\ -\ 25x\ +\ 11\ =\ 0]


Done.



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