Question 181648
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You <i><b>might</b></i> be correct.  It all depends on what you meant by 3x/9x-15.


If you meant:


*[tex \LARGE \text{          }\math \frac{3x}{9x} - 15]


Then you are absolutely correct.


On the other hand, if you meant:


*[tex \LARGE \text{          }\math \frac{3x}{9x-15}]


Then you have a bit of a problem with your process.  You cannot separate the terms in a polynomial numerator or denominator and then operate with the separated terms individually.  Let me demonstrate on this problem with a couple of simple substitutions:


Let <i>x</i> = 1, then 


*[tex \LARGE \text{          }\math \frac{3(1)}{9(1)- 15}  = \frac {3}{-6} = -\frac {1}{2} \neq -\frac {44}{3}]


or let <i>x</i> = 2, then 


*[tex \LARGE \text{          }\math \frac{3(2)}{9(2)- 15}  = \frac {6}{3} = 2\neq -\frac {44}{3}]


What you can do with this problem is factor out the 3 that is common to both the numerator and both terms of the denominator and then eliminate it, thus:


*[tex \LARGE \text{          }\math \frac{3(x)}{3(3x-5)} = \left(\frac{3}{3}\right)\left(\frac{x}{3x-5}\right) = 1 \ast \left( \frac{x}{3x-5} \right) = \frac{x}{3x-5}]


And that is as much as you can do with this expression, except to make sure you restrict the value of <i>x</i> such that *[tex \Large x \neq \frac {5}{3}] because that value would make the denominator equal 0.


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