Suppose he originally bought N pigs for $150, so he paid $150/N per pig.  Then 8
died, so he then had only N-8 pigs to sell.  He then sold N-8 pigs each for $6
more than he paid, or $150/N + 6 each.  So he took in $[150/N + 6][N-8].  This was $92
more than the $150 he paid, or $150+$92 or $242.  So the equation is

$[150/N + 6][N-8] = $242

[150/N + 6][N-8] = 242

Clear the fraction by multiplying both sides by N

N∙[150/N + 6][N-8] = N∙242

[150 + 6N][N-8] = 242N

150N-1200+6N²-48N = 242N

6N²+102N-1200 = 242N 

6N²-140N-1200 = 0

Divide all terms on both sides by 2

3N²-70N-600 = 0

(3N+20)(N-30) = 0

3N+20=0; N-30=0

3N=-20;   N=30

    N=-20/3

Ignore the negative fractional answer.

Solution: 30 pigs

Let's check:

He bought 30 pigs for $150. So he paid $150/30 = $5 per pig.  Then 8 died.  So
he only had 22 pigs to sell. He sold each of the 22 pigs for $6 more than he
paid, so he sold each pig for $11. Since he sold 22 pigs, he ended up with
22∙$11 = $242, and since he paid $150 for the pigs in the beginning, his profit
was $242-$150 = $92.

Edwin