Combination of three animals

We start out with systems of equations. Let $g=\text{ number of goats bought}$ , $c=\text{ number of cows bought}$ , and $h=\text{ number of horses bought}$ . Since each goat costs 50 cents and 100 cents = 1 peso, each goat costs $\frac{50}{100}=\frac{1}{2}$ pesos, and we have

$\frac{g}{2}+c+3h=10$

$g+c+h=10$ .

If we subtract the 1st equation from the second equation, we get $-\frac{g}{2}+2h=0$ , or $2h=\frac{g}{2}$ , or $g=4h$ . Since he bought at least 1 of each animal, he couldn't have bought 0 horses and goats, and if he bought 2 horses, then he'd buy 8 goats, totaling 10 animals, and he didn't buy any cows. Therefore, he bought 1 horse, $\boxed{4}$ goats, and 5 cows.