Question 203563
When solving a rational equation, why is it necessary to perform a check?
<pre><font size = 4 color = "indigo"><b>
In solving a rational equation, we always clear of fractions.

When we clear of fractions we always multiply both sides by a
variable quantity.

When we multiply both sides of an equation by a variable expression,
we may introduce an extraneous solution a zero of that expression 
we multiplied through by.   

Example:

{{{2/(x+2)+1/(x-3)= -10/((x+2)(x-3))}}}

The LCD is {{{(x+2)(x-3)}}}
  
So we multiply through by {{{((x+2)(x-3))/1}}}

{{{
(((x+2)(x-3))/1)(2/(x+2))

+

(((x+2)(x-3))/1)(1/(x-3))
=
 (((x+2)(x-3))/1)(-10/((x+2)(x-3)))

}}}

{{{
(((cross(x+2))(x-3))/1)(2/(cross(x+2)))

+

(((x+2)(cross(x-3)))/1)(1/(cross(x-3)))
=
 (((cross(x+2))(cross(x-3)))/1)(-10/((cross(x+2))(cross(x-3))))

}}}

{{{2(x-3)+(x+2)=-10}}}

{{{2x-6+x+2=-10}}}

{{{3x-4=-10}}}

{{{3x=-6}}}

{{{x=-6/3}}}

{{{x=-2}}}

So at first it appears that {{{x=-2}}} is
a solution.  However when we check:

{{{2/(x+2)+1/(x-3)= -10/((x+2)(x-3))}}}

{{{2/((-2)+2)+1/((-2)-3)= -10/(((-2)+2)((-2)-3))}}}

{{{2/0+1/(-5)= -10/(0)(-5))}}}

{{{2/0+1/(-5)= -10/0)}}}

But there are zeros in two denominators!

So there is no solution to this rational equation.

Edwin</pre>