Question 446771
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(y^2-4/y+3) = 2-(y-2/y+3)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~



As it is written in the post, it is inaccurate writing and it does not correspond to real intent.


The correct equation is


        {{{(y^2-4)/(y+3)}}} = {{{2-(y-2)/(y+3)}}}.



Now, @mananth gives the solutions   y = 4  OR  y = -3.


But if you will present it in this form to your teacher,  you can easily get the score of  '2'.


It is because  y = -3   IS  NOT   a solution.   It is not even in the domain of this equation.


Therefore,  the solution to this problem should start with these words


<pre>
     The domain of this equation is the set of all real numbers except of &nbsp;y = -3, 
     because &nbsp;(y+3) &nbsp;is in the denominator of the formulas.
</pre>

After that, you make all this analysis, &nbsp;which @mananth makes, &nbsp;but at the end you say


<pre>
    Since the root y = -3 is out of the domain, the only solution to the given equation is y = 4.
</pre>

It is how to present the complete solution correctly and to get the highest score.