## Task A-1.3.

Farmer Ivan on his farm has chickens, pigs, and sheep. His animals have a total of 46 heads and 124 legs. If he were to double the number of chickens and triple the number of sheep on the farm, with the same number of pigs, the total number of legs of all the animals on the farm would be 232. How many heads would there be in this case?
## Solution.

Let $K$ be the number of chickens, and $N$ be the number of other (four-legged) animals. From the first two conditions of the problem, we get

$$
\begin{aligned}
K+N & =46 \\
2 K+4 N & =124
\end{aligned}
$$

Substituting $N=46-K$ from the first equation into the second, we get

$$
\begin{aligned}
2 K+4 \cdot(46-K) & =124 \\
184-2 K & =124 \\
K & =30
\end{aligned}
$$

Let $2 K$ and $N^{\prime}$ be the numbers of animals in the second case. From the condition on the number of legs, we get

$$
232=2 \cdot 2 K+4 N^{\prime}=120+4 N^{\prime}
$$

from which we get $N^{\prime}=28$.

In this case, the number of animal heads on the farm would be

$$
2 K+N^{\prime}=2 \cdot 30+28=88
$$

Note: The problem can be solved by a system of 3 equations with 3 unknowns:

$$
\begin{aligned}
K+S+O & =46 \\
2 K+4 S+4 O & =124 \\
2 \cdot 2 K+4 S+4 \cdot 3 O & =232
\end{aligned}
$$

where $K$ is the number of chickens, $S$ is the number of pigs, and $O$ is the number of sheep on the farm. Setting up the system earns 1 point, determining one of the values of the unknowns $(O=6, S=10, K=30)$ earns 2 points, determining the remaining two values of the unknowns earns 1 point each, and finally determining the new number of heads by the equation

$$
2 K+S+3 O=2 \cdot 30+10+3 \cdot 6=88
$$

earns 1 point.