Question 1026531
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4/(x-3) + (2x)/(x^2-9) = 1/(x+3)
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        This problem is very tricky.

        It contains a hidden trap that's easy to fall into.

        Tutor @mananth didn't avoid this trap — he fell into it and produced an incorrect solution, 

        for which he deserves a solid F.


        Below is my correct solution to replace @mananth's incorrect one.



<pre>
Your starting equation is 

    {{{4/(x-3)}}} + {{{(2x)/(x^2-9)}}} = {{{1/(x+3)}}}.


Notice that the domain of this equation is the set of all real numbers except of x = 3 and/or x = -3,
where the denominators are not defined.


So, we will work on the domain x =/= -3  and  x =/= 3.


Multiply both sides by  (x^2-3).  You will get 

    4(x+3) + 2x = x - 3,

    4x + 12 + 2x = x - 3,

    6x - x = -3 - 12

       5x  =    -15,

        x  =    -15/5 = -3.


But x = -3 is not in the domain.


So, we conclude that the given equation HAS NO solutions in its domain.    <<<---===  <U>ANSWER</U>
</pre>

Solved correctly.