There are 60 animals in a yard. Some are chickens, some are dogs. Together, they have 194 legs. How many chickens and how many dogs are there?

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In this problem, we have a yard with 60 animals, which are a mix of chickens and dogs. Together, these animals have a total of 194 legs. We need to find out how many chickens and how many dogs are in the yard. Let's set up a system of equations to solve this. If we let C represent the number of chickens and D represent the number of dogs, we can write our first equation as C plus D equals 60, which represents the total number of animals. For our second equation, we need to account for the total number of legs. Since each chicken has 2 legs and each dog has 4 legs, we can write 2C plus 4D equals 194. Now, let's solve this system of equations. First, we'll express C in terms of D using our first equation. From C plus D equals 60, we get C equals 60 minus D. Next, we substitute this expression for C into our second equation. So, 2 times (60 minus D) plus 4D equals 194. Let's expand this: 120 minus 2D plus 4D equals 194. Simplifying the left side: 120 plus 2D equals 194. Rearranging to isolate the term with D: 2D equals 194 minus 120, which gives us 2D equals 74. Therefore, D equals 37. This means we have 37 dogs in the yard. Now that we know there are 37 dogs, we can find the number of chickens by substituting this value back into our first equation. We have C equals 60 minus D, so C equals 60 minus 37, which gives us C equals 23. This means there are 23 chickens in the yard. Let's verify our solution by checking both equations. For the first equation, C plus D equals 23 plus 37, which equals 60. This matches our given total number of animals. For the second equation, 2C plus 4D equals 2 times 23 plus 4 times 37, which equals 46 plus 148, for a total of 194 legs. This matches our given total number of legs. Therefore, our solution is correct: there are 23 chickens and 37 dogs in the yard. Let's explore an alternative method to solve this problem using elimination. We start with our system of equations: C plus D equals 60, and 2C plus 4D equals 194. First, we multiply the first equation by 2 to get 2C plus 2D equals 120. Now, we can subtract this from the second equation to eliminate the C terms. From 2C plus 4D equals 194, we subtract 2C plus 2D equals 120. This gives us 2D equals 74, so D equals 37. Now we substitute this value back into our first equation: C plus 37 equals 60, which gives us C equals 23. This confirms our previous result: there are 23 chickens and 37 dogs in the yard. The elimination method provides a systematic approach to solving systems of linear equations without having to substitute expressions. Let's summarize what we've learned. We started with a problem involving 60 animals, which were a mix of chickens and dogs, with a total of 194 legs. We set up a system of two equations: C plus D equals 60, representing the total number of animals, and 2C plus 4D equals 194, representing the total number of legs. Using the substitution method, we found that D equals 37 and C equals 23. We verified our answer by checking that 23 chickens with 46 legs plus 37 dogs with 148 legs indeed gives us a total of 194 legs. We also demonstrated an alternative approach using the elimination method, which gave us the same answer. This problem illustrates how systems of linear equations can be used to solve real-world problems involving multiple unknowns.