Question 23194
<pre>under waht situation would one or more solutions of a rational equation be
unacceptable?
<b><font size = 3>
For rational equations, extraneous solutions are values that cause
any denominator in the original problem to be 0.  Of course, when we
have 0 in the denominator we have an expression that is undefined.
So, we would have to discard any values that would cause the
denominator to be 0.  

Example: Solve for x

  3       x      3
----- = ----- - ---
 x-3     x-3     2

 Step 1: Simplify by removing the fractions. 
 Mult. both sides by LCD of 2(x-3) 

         3             x            3
2(x-3)·----- = 2(x-3)----- - 2(x-3)---
        x-3           x-3           2
  
Cancel where possible:

   <sub>1</sub>     3        <sub>1</sub>    x      <sub>1</sub>      3
2<s>(x-3)</s>·----- = 2<s>(x-3)</s>----- - <s>2</s>(x-3)---
        <s>x-3</s>           <s>x-3</s>           <s>2</s>
         <sup>1</sup>             <sup>1</sup>             <sup>1</sup>


Step 2: Solve the remaining equation. 

            6 = 2x - 3(x-3)   *Remove ( ) by using dist. prop.
            6 = 2x - 3x + 9   *Inverse of add. 9 is sub. 9 
            6 = -x + 9
        6 - 9 = -x + 9 - 9
           -3 = -x

          -3     -x
         ---- = ----          *Inverse of mult. by -1 is div. by -1
          -1     -1
       
            3 = x
  
Step 3: Check for extraneous solutions.  

Note that 3 does cause two of the denominators of the original equation
to be zero.  So 3 is an extraneous solution.  That means there is no
solution. 

The answer is: NO solution.
 
Edwin
AnlytcPhil@aol.com</pre>