Question 333521
Wne solving rational equations, we always have to clear of fractions
which involves multiplying through by an LCD which is a variable expression.
But it is possible that that variable LCD that we multiplied through by
could equal zero.  If that were the case then we would have made a violation
of multiplying through an equation by zero. 

Take this case:

{{{3/(x-7)}}}{{{""+""}}}{{{1/(x-3)}}}{{{""=""}}}{{{12/((x-7)(x-3))}}}

If we multiply through by the LCD = {{{(x-7)(x-3)}}}, we are assuming
that x is not equal to either 7 or 3.  For it x were equal to either
7 or 3 then two of those denominators would be 0. 

So we go ahead and multiply through by the LCD and get:

{{{3(x-3)+x-7 = 12}}}
{{{3x-9+x-7=12}}}
{{{4x-16=12}}}
{{{4x=28}}}
{{{x=7}}}

But in this case x would have to have equaled to 7.
That means we were in violation of multiplying through by 0.
So we got an extraneous solution, which is not a solution at all.

We discover this by substituting in the original equation:

{{{3/(x-7)}}}{{{""+""}}}{{{1/(x-3)}}}{{{""=""}}}{{{12/((x-7)(x-3))}}}

{{{3/(7-7)}}}{{{""+""}}}{{{1/(7-3)}}}{{{""=""}}}{{{12/((7-7)(7-3))}}} 

{{{3/0}}}{{{""+""}}}{{{1/4}}}{{{""=""}}}{{{12/((0)(4))}}}
 
{{{3/0}}}{{{""+""}}}{{{1/4}}}{{{""=""}}}{{{12/0}}}

That is meaningless because 0 in the denominator is not defined.

Edwin</pre>