Question 72722
<pre>

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)
                     {{{60n - 20 - 12n - 4 = (27/2)n^2 - 3/2}}}
   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