Solve the problem 

     20        4

    ____  -   ____  = 3/2

    3n+1      3n-1



Find the LCD of 3n + 1 nd 3n - 1

 LCD = (3n + 1)(3n - 1)

Multiply the LCD both sides



                      20        4

[(3n + 1)(3n - 1)]   ____  -   ____  = 3/2  [(3n + 1)(3n - 1)]

                     3n+1      3n-1



      20 (3n - 1) - 4(3n + 1) = 3/2 (3n + 1)(3n - 1)



Simplify:

     [20(3n) - 20 (1)] - [4(3n) + 4(1)] = 3/2(9n^2 - 1)

                     

   48n - 24 = (27/2)n^2 - 3/2

   

Multiply 2 both sides



   96n - 48 = 27n^2 - 3



          0 = 27n^2 - 96n + 45

            

Factor

          0 = (9n - 5)(3n - 9)     



Apply zero product property



      9n - 5 = 0       and      3n - 9 = 0

         9n  = 5                    3n = 9

          n  = 5/9                   n = 3





If you want to see if the values of your n are correct, 

substitute n = 5/9 and n = 3 to the equation.         





Checking 



     20        4

    ____  -   ____  = 3/2

    3n+1      3n-1



When n = 5/9



        20            4

    ________  -   ________  = 3/2

    3(5/9)+1      3(5/9)-1





      20         4

    _____  -   ______  = 3/2

    5/3+1      5/3-1



     20        4

    ____  -   ____  = 3/2

     8/3       2/3 





    20*3       4*3

   ______  -  _____  = 3/2

     8          2



      60      12      3

    _____  - ___  = ____  

      8       2       2



     30         3

    ____ - 6 = ___

     4          2



     30     24    3

    ____ - ___ = ___

     4      4     2



        6      3

       ___  = ___

        4      2



       3/2 = 3/2  ------> True



When n = 3



      20        4

    ____  -   ______  = 3/2

    3(3)+1    3(3)-1





      20        4

    ____  -   ______  = 3/2

    9 + 1      9 -1





     20         4

    ____  -   ______  = 3/2

     10         8



     2 - 1/2  = 3/2



    4       1

   ___  -  ___ = 3/2

    2       2



      3/2 = 3/2   ------> True







Therefore the solutions are n = 5/9 and n = 3 

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