Three countrymen met at a cattle market. "Look here," said Hodge to Jakes, "I'll give you six of my pigs for one of your horses, and then you'll have twice as many animals here as I've got."

"If that's your way of doing business," said Durrant to Hodge, "I'll give you fourteen of my sheep for a horse, and then you'll have three times as many animals as I."

"Well, I'll go better than that," said Jakes to Durrant; "I'll give you four cows for a horse, and then you'll have six times as many animals as I've got here."

No doubt this was a very primitive way of bartering animals, but it is an interesting little puzzle to discover just how many animals Jakes, Hodge, and Durrant must have taken to the cattle market.

**Answer:**

Jakes must have taken 7 animals to market, Hodge must have taken 11, and Durrant must have taken 21. There were thus 39 animals altogether.

This can be solved with a little algebra, where J - Jakes, H = Hodge and D = Durrant:

From the clues:

2 x (H - 6 + 1) = J + 6 - 1 [1]

3 x (D - 14 + 1) = H + 14 - 1 [2]

6 x (J -4 + 1) = D + 4 - 1 [3]

These can be rearranged to give:

2H - J = 15 [4]

3D - H = 52 [5]

6J - D = 21 [6]

We can now use 2 x [5] + [4] to give:

6D - J = 119 [7]

We can now use [7] + 6 x [6] to give:

35J = 245

J = 7

We can then use J = 7 in [4] and [6] to give H = 11, and D = 21.