Question 274297
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Then you have an extraordinarily difficult task ahead of you, one for which I cannot provide any assistance nor can anyone else.  The problem is you cannot "solve"


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{2xy\ +\ 5x\ -\ 2y\ -\ 5}{3xy\ +\ 4x\ -\ 3y\ -\ 4}]


You cannot solve anything that does not have an equals sign or some sort of relational operator.


It is a rather complex rational expression that can be <b><i>simplified</i></b>.  If that is what you meant, then follow along:


Start with the numerator polynomial.  Factor an *[tex \Large x] out of the first two terms and factor a -1 out of the last two terms:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{x\left(2y\ +\ 5\right)\ -\ 1\left(2y\ +\ 5\right)}{3xy\ +\ 4x\ -\ 3y\ -\ 4}]


Next, factor *[tex \Large 2y\ +\ 5] from the two terms that remain:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\left(x\ -\ 1\right)\left(2y\ +\ 5\right)}{3xy\ +\ 4x\ -\ 3y\ -\ 4}]


Next work on the denominator.  Use a similar strategy, factoring *[tex \Large x] out of the first two terms and -1 out of the last two terms:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\left(x\ -\ 1\right)\left(2y\ +\ 5\right)}{x\left(3y\ +\ 4\right)\ -\ 1\left(3y\ +\ 4\right)}]


And then factoring *[tex \Large 3y\ +\ 4] from the two terms that remain:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\left(x\ -\ 1\right)\left(2y\ +\ 5\right)}{\left(x\ -\ 1\right)\left(3y\ +\ 4\right)}]


Eliminate the factor of *[tex \Large x\ -\ 1] that is common to both numerator and denominator leaving you with:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{2y\ +\ 5}{3y\ +\ 4}]



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