Rational Equations

Date: 06/07/97 at 23:06:53
From: Catherine McDonald
Subject: Polynomials

I'm in 9th grade and I have a final coming up. I thought I'd try to 
understand how to solve rational equations before I get to the final.  

For something like (1/x+1) - (1/x+2) = 1/2, I understand the cross 
multiplying part, but then what do you do? One teacher said that you 
had to find the common multiple, but then another teacher said 
something different.  

Please try to explain this problem. Another example would be 
fantastic.

Thanks in advance,
Catherine

Date: 06/08/97 at 10:52:22
From: Doctor Anthony
Subject: Re: Polynomials

Dear Catherine,

Be careful how you write down the fractions because the method you use 
could be interpreted as 1/x + 1 - 1/x - 2, which I am sure you did not 
intend.

We have: 1/(x+1) - 1/(x+2) = 1/2

To add fractions you must have the same denominator in each fraction.  
An example from arithmetic would be: 1/2 + 1/3.  The common 
denominator will have to be a number into which both 2 and 3 will 
divide. Obviously 6 is such a number.  To make the denominator 6 in 
the fraction 1/2, multiply top and bottom by 3 to get 3/6.  To make 
the denominator 6 in the fraction 1/3, multiply top and bottom by 2 to 
get 2/6.  We now have to add 3/6 + 2/6, and since the denominators are 
the same we can add the numerators to get:

  (3 + 2)/6 =  5/6  

On paper you would write this:

        3  +  2        5
       ---------  =  -----
           6           6

Looking at the algebra problem, the common denominator will be the 
product (x+1)(x+2). Since we multiply the denominator of the first 
fraction by (x+2), we must multiply the numerator also by (x+2).

The first fraction could be written:       x+2
                                        ----------
                                        (x+1)(x+2)

The second fraction could be written:     x+1
                                       ----------
                                       (x+1)(x+2)

In practice you would write this:  

           (x+2) - (x+1)            1
           --------------  =   -----------
             (x+1)(x+2)         (x+1)(x+2)

Now we can start to solve the equation:

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


Cross multiplying:    2 = (x+1)(x+2)
                      2 = x^2 + 3x + 2
                      0 = x^2 + 3x
                      0 = x(x+3)

So we have two possible values of x which satisfy the equation; either 
x = 0 or x = -3. We can check if these are correct:

If x = 0, the equation becomes 1/1 - 1/2 = 1/2, which is correct.

If x = -3, the equation becomes 1/-2 - 1/-1 = -1/2 + 1 = 1/2, again 
correct.  

-Doctor Anthony,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/