Welcome to the classic Chickens and Rabbits problem! This is an ancient Chinese mathematical puzzle. Imagine we have some chickens and rabbits together in a cage. We can count the total number of heads and the total number of feet, but we need to figure out exactly how many chickens and how many rabbits there are.
Let's set up this problem mathematically. We'll use x to represent the number of chickens and y to represent the number of rabbits. For this example, let's say we have 8 total heads and 22 total feet. Since each animal has exactly one head, we get our first equation: x plus y equals 8. For the feet, chickens have 2 feet each and rabbits have 4 feet each, giving us our second equation: 2x plus 4y equals 22.
Now let's solve this system of equations using substitution. From the first equation x plus y equals 8, we can solve for x to get x equals 8 minus y. Next, we substitute this into the second equation: 2 times the quantity 8 minus y, plus 4y equals 22. Expanding this gives us 16 minus 2y plus 4y equals 22. Simplifying, we get 16 plus 2y equals 22. Solving for y: 2y equals 6, so y equals 3. Finally, substituting back: x equals 8 minus 3, which equals 5.
Let's verify our solution. We found 5 chickens and 3 rabbits. For heads: 5 plus 3 equals 8, which matches our given total. For feet: 5 chickens times 2 feet each gives 10 feet, plus 3 rabbits times 4 feet each gives 12 feet, for a total of 22 feet. This also matches our given total. Both conditions are satisfied, so our answer is correct: there are 5 chickens and 3 rabbits in the cage.
To summarize what we've learned: The Chickens and Rabbits problem is a classic way to introduce systems of equations. We set up equations based on the given constraints, use algebraic methods like substitution to solve them, and always verify our answers. This problem-solving approach can be applied to many similar counting and constraint problems in mathematics and real life.