Counting Sheep

If we denote David by A, Harry by B, Sam by C and Peter by D we can find the answer in the following way. Since A + B + C = D and A + B + C + D = 44, it is easy to see that D has 22 sheep. So A + B + C = 22. We can use this along with the two other facts, A + 3 = B - 1 and C - 3 = 3A, to find that A has 3, B has 7, C has 12 and D has 22. In a little more detail:

A + B + C = 22 (1) A + 3 = B - 1 (2) C - 3 = 3A (3)

Rearrange (3) to give C = 3A + 3 and replace the C in (1) to give A + B + 3A + 3 = 22. Simplified this gives B = 19 - 4A (4). Rearrange (2) to give B - A = 4. Use (4) in the place of B to give 19 - 4A - A = 4 to give A = 3. Now use A to work out B, and then C.

$\begin{aligned} \text {Let David have "w" sheep.}\\ \text {Let Harry have "x" sheep.}\\ \text {Let Sam have "y" sheep.}\\ \text {Let Peter have "z" sheep.}\\ \\ \text{From first condition}:-\\ w + 3 & = x - 1\\ w & = x - 4 \longrightarrow \boxed{1}\\ \\ y & = w + x + z\\ w & = y - x - z \longrightarrow \boxed{2}\\ \\ z - 3 & = 3w\\ 3w & = z - 3 \longrightarrow \boxed{3}\\ \\ w + x + y + z & = 4 × 11 = 44 \longrightarrow \boxed{4}\\ \\ \\ \text{Adding 1 and 2, we get}:-\\ 2w & = y - z - 4\\ z & = y - 4 - 2w \longrightarrow \boxed{5}\\ \\ \text{Adding 5 and 3, we get}:-\\ 5w & = y - 7 \longrightarrow \boxed{6}\\ \\ \\ \text{ From 1}:-\\ x & = w + 4\\ \\ \text{From 5}:-\\ z & = y - 4 - 2w\\ \\ \\ \text{Substituting the above two equations in 4, we get}:-\\ w + w + 4 + y + y - 4 - 2w & = 44\\ 2y & = 44\\ y & = 22\\ \\ \\ \text{Substituting y = 22 in 6, we get}:-\\ 5w & = 22 - 7\\ 5w & = 15\\ w & = \boxed {3}\\ \\ \text{So, David has 3 sheep} \end{aligned}$