Welcome to the classic chicken and rabbit problem! We have a cage with chickens and rabbits. There are 10 heads and 26 legs in total. Each chicken has 1 head and 2 legs, while each rabbit has 1 head and 4 legs. Our goal is to find how many chickens and how many rabbits are in the cage.
To solve this problem systematically, let's define variables. Let C represent the number of chickens and R represent the number of rabbits. We know that each chicken has 1 head and 2 legs, while each rabbit has 1 head and 4 legs. This gives us two equations: C plus R equals 10 for the total heads, and 2C plus 4R equals 26 for the total legs.
Now let's solve the system of equations step by step. From the first equation C plus R equals 10, we can express C as 10 minus R. Next, we substitute this into the second equation: 2 times open parenthesis 10 minus R close parenthesis plus 4R equals 26. Expanding this gives us 20 minus 2R plus 4R equals 26, which simplifies to 20 plus 2R equals 26. Solving for R, we get 2R equals 6, so R equals 3. Finally, substituting back, C equals 10 minus 3, which equals 7.
Let's verify our solution. We found 7 chickens and 3 rabbits. Checking the heads: 7 plus 3 equals 10 heads, which matches our given information. Checking the legs: 7 chickens times 2 legs each equals 14 legs, plus 3 rabbits times 4 legs each equals 12 legs, for a total of 26 legs. This also matches perfectly! Therefore, our answer is correct: there are 7 chickens and 3 rabbits in the cage.
To summarize what we've learned: We successfully solved the classic chicken and rabbit problem using algebraic methods. We set up a system of two equations based on the constraints of heads and legs, used substitution to find that there are 7 chickens and 3 rabbits, and verified our solution. This systematic approach can be applied to solve many similar counting problems in mathematics.