Question 164064
<pre><font size = 4 color = "indigo"><b>
{{{5/(y-3) = (y+7)/(2y-6) + 1}}}

Write the {{{1}}} as {{{1/1}}} so every term will be
a fraction:

{{{5/(y-3) = (y+7)/(2y-6) + 1/1}}}

Factor {{{2y-6}}} by taking out a {{{2}}}, as {{{2(y-3)}}}

{{{5/(y-3) = (y+7)/(2(y-3)) + 1/1}}}

Put every term in parentheses:

{{{(5/(y-3)) = ((y+7)/(2(y-3))) + (1/1)}}}

The LCD is {{{2(y-3)}}}.  Put it over {{{1}}} and in
parentheses as {{{((2(y-3))/1)}}} and multiply through by it:

{{{(2(y-3)/1)(5/(y-3)) = (2(y-3)/1)((y+7)/(2(y-3))) + (2(y-3)/1)(1/1)}}}

Cancel what will cancel:

{{{(2(cross(y-3))/1)(5/(cross(y-3))) = (cross(2)(cross(y-3))/1)((y+7)/(cross(2)(cross(y-3)))) + (2(y-3)/1)(1/1)}}}

All that's left is

{{{2*5=y+7+2(y-3)}}}

{{{10=y+7+2y-6}}}

{{{10=3y+1}}}

{{{9=3y}}}

{{{3=y}}}

Check to see if y=3 is a solution:

{{{5/(y-3) = (y+7)/(2y-6) + 1}}}

{{{5/(3-3) = (y+7)/(2(3)-6) + 1}}}

{{{5/0 = (y+7)/(6-6)+1}}}

{{{5/0 = (y+7)/0 + 1}}}

That contains division by 0 and that is 
always undefined.  So {{{y=3}}} is not a
solution.  It is call an "extraneous 
solution", and not a genuine solution.
Therefore there is no solution to the
problem.

Edwin</pre>