How do I solve equations containing fractions ?

Question

x/x+2  -   x/x-2=  x+20/x^2-4

 

Answer 1

Multiply it out so that each denominator shares the same expressions. Since, (x+2)*(x-2)=x²-4 you can multiply the first term by (x-2)/(x-2) and the second term by (x+2)/(x+2) so that all terms have the same denominator. This does not change the value of each term since multiplying the same value to the numerator and denominator is the same things as multiplying the term by 1 (for example, (x-2)/(x-2)=1). Once all of the terms have the same denominators, the numerators can be added together and put over the same denominator.

 

x/x+2 - x/x-2 = x+20/x²-4

x*(x-2) / (x+2)*(x-2) -  x*(x+2) / (x-2)*(x+2) = x+20 / x²-4

(x²-2x / x²-4) - (x²+2x / x²-4) = x+20 / x²-4

(x²-2x / x²-4) - (x²+2x / x²-4) - (x+20 / x²-4) = 0

[(x²-2x) - (x²+2x) - (x+20)] / (x²-4) = 0

[x²-2x - x² - 2x - x - 20)] / (x² - 4) = 0

(-5x - 20) / (x² - 4) = 0

-->  Either (-5x-20)=0 or (x²-4)=0. However you cannot divide by a negative number, so (x²-4) cannot equal 0.

-->  -5x - 20 = 0

-5x = 20

x = 20 / -5 = -4

Answer 2

   x/(x+2) - x/(x-2) = (x+20)/(x²-4)

 

     x/(x+2) - x/(x-2) = (x+20) / ((x+2)(x-2))     

         [Once you recognize the difference of squares, this problem is easy]

 

     (x-2)x   - (x+2)x   = (x+20)         [multiply all by ((x+2)(x-2)),

                                                       the LCM of the denominators]

       x² - 2x   - x² - 2x   =  x + 20

                   -4x =  x + 20

                    -5x = 20

                  -4 = x

 

Checking (very important):

  Is     (-4)/(-4+2)  -  (-4)/(-4-2) = (-4+20)/((-4)²-4)   ?

                4/2      -  4/6         =   16/12   ?

                 24/12   -  8/12   =   16/12   ?

                        16/12 = 16/12   ?yes