Question 299158
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Solve by the LCM method:

{{{6/(y+3)}}} + {{{2/y}}} = {{{(5y-3)/(y^2-9)}}}.

I am unable to come up with the LCD:  (y+3)(y-3)?  Unsure of how to work this. 
Once I have the LCD, I believe I will be able to solve it.  Thank you in advance for your help.  
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        The solution in the post by @mananth is incorrect.

        I came to bring a correct solution.



<pre>
LCM (the Least Common Multiple) is y*(y^2-9) = y*((-3)*(y+3).

So, we multiply both sides of the given equation by this expression y*(y-3)*(y+3).
We get

    6y*(y-3) + 2(Y^2-9) = (5y-3)y,

    6y^2 - 18y + 2y^2 - 18 = 5y^2 -3y,

    3y^2 - 15y - 18 = 0,

    y^2 - 5y - 6 = 0,

    (y-6)*(y+1) = 0.


The roots are  y = 6  and/or  y = -1.


<U>ANSWER</U>.  The solutions for the given equation are  y= 6  and/or  y = -1.


To  {{{highlight(highlight(check))}}},  I used a free of charge plotting tool at web-site  http:\\www.desmos.com/calculator/


I printed the formulas for the left side and right side functions and got two plots.

They have exactly two intersection points at y = 6  and  y = -1  , confirming my solution.
</pre>

Solved correctly.