Solution:
Given: S=n/2[2a+(n-1)d] ...............(1)

Equation (1) can be written as,

             S=na + n(n-1)d/2

===> S-n(n-1)d/2 =na

===> S/n -(n-1)d/2 =a         ( divide both sides by n.)

===> a= [2S - n(n-1)d]/2n

Answer: a= [2S - n(n-1)d]/2n

we have
S= n/2 [ 2a+ (n-1)d ]
take the cross product

2S = n[ 2a+ (n-1)d ]

2S   = 2an + n(n-1)d                 ( using left distributive law)

2S - n(n-1)d = 2an

2an = 2S - n(n-1)d

 a =   1   [2S - n(n-1)d]
         2n

Question:
rearrange S= n/2 [ 2a+ (n-1)d ] to get a. please show working

get a on its own
S= n/2 [ 2a+  (n-1)d ].

Solution:
(Two  of our MindZinger experts answered. I agree with both for their contribution.)

Please think that you are given  the values  of S, n,   d  and  this equation   which relates 
S,n ,a and  d . Now you have to find the  value of  a  .
Here we treat  a , as  an unknown and  find its value  through the other known values of S,n and d.
The method of finding a  is quite well expressed  already and  I repeat below :

S = (n/2)[2 a +(n-1)d],  Given. divide both sides by n/2 to get,
S*(2/n)  =  2a + (n-1)d. Subtract (n-1)d from both sides to get,
(2S/n ) -(n-1)d  = 2 a.  Thus,
a  = (S/n) - (n-1)d/2.