Question 191061
<font face="Garamond" size="+2">


Well, I will give you high marks for creativity.  But your answer isn't even close.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \frac{3x}{x^2 - 1} = \frac{5}{x}]


The two denominators have no factors in common, so the LCD is simply the product of the two, or *[tex \LARGE x(x^2 - 1)].  Multiplying that out would only confuse the issue, so let's leave it the way it is.  Applying the LCD:



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{3x}{(x^2 - 1)}\left(\frac{x}{x}\right) = \frac{5}{x}\left(\frac{x^2-1}{x^2-1}\right) \ \ \Rightarrow\ \  \frac{3x^2}{x(x^2 - 1)} = \frac{5(x^2-1)}{x(x^2 -1)} ]


Now, use the distributive property on the numerator on the right:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \frac{3x^2}{x(x^2 - 1)} = \frac{5x^2-5}{x(x^2 -1)}]


Now we have two fractions equal to each other and they have equal denominators.  Therefore the numerators must be equal as well.  So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3x^2 = 5x^2 - 5]


Add *[tex \LARGE -5x^2] to both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -2x^2 = -5]


Multiply both sides by *[tex \LARGE -{1 \over 2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2 = \frac{5}{2}]


Take the square root:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = \pm sqrt{\frac{5}{2}} = \pm \frac{sqrt{5}}{sqrt{2}}]


But we still need to rationalize the denominator so multiply by 1 in the form of *[tex \LARGE \frac{sqrt{2}}{sqrt{2}}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = \pm \left(\frac{sqrt{5}}{sqrt{2}}\right) \left(\frac{sqrt{2}}{sqrt{2}}\right) = \pm \frac{sqrt{10}}{2}]


You should check both answers to make certain that we didn't create an extraneous root in the action of squaring the variable during the solving process.


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