Time and Works: 
A can built a wall in 8 days while b can destroy the same wall in 3
  days. A start the work and worked for 6 days, during the last 2 days of
  which B also joins him. In how many days A finish the remaining work
  alone

I have tried :
I have formed one equation 
6/8 - 2/3 + x =1


This equation states that A works for 6 days out of 8 days and b
  destroys the the wall so i subtracted the equation and the remaining
  work i placed as x which is equal full part work 1

That equation showing wrong answer,what I am doing mistake, please guide me answer

with B joins A to destroy the wall , guide me answer
 A: Your equation should rather be:
6/8 - 2/3 + x/8 =1

It means
$$
t_1 r_a + t_2 r_b + t r_a = 1 \,\text{wall}
$$
with given times
$$
t_1 = 6 \,\text{days} \\
t_2 = 2 \,\text{days} \\
$$
unknown time $t$
and rates
$$
r_a = \frac{1 \,\text{wall}}{8\,\text{days}} \\
r_b = - \frac{1 \,\text{wall}}{3\,\text{days}} \\
$$
A: It's all about the unitary method : convert each person's work to "work per day".
$A$ makes the wall in $8$ days, so in one day, he does $\frac 18$ work.
$B$ destroys the wall in three days, so in  one day, he does $-\frac 13$ work (remember, destruction of the wall is negative work in our context).
Hence, it is mentioned that $A$ worked for six days. Four of these days he worked separately,that means he completed $4 \times \frac 18 = \frac 48$ of the work.
Then $B$ joined him. $B$ and $A$ together, would do $2\left(\frac 13 - \frac 18\right) = -\frac {10}{24}$ amount of work. 
That is, at the end of six days, the total amount of work done is $\frac 48 - \frac{10}{24} = \frac{1}{12}$ (after simplification)
So the remaining work is $1-\frac{1}{12} = \frac{11}{12}$. The time taken by $A$ to do this is $\frac{11}{12} \div \frac 18 = \frac{22}{3}$ days. Hence, $A$ requires $22$ days and $16$ hours more, to finish his work.
Now, if $B$ takes $3$ days to build a wall rather than destroy it, then his contribution per day will be $+ \frac 13$, not $-$. I leave you to see the situation here (just tweak the suitable equation).    
