Here is a problem I encountered in an algebra class many years ago: A certain barn has horses and chickens, 50 animals in all. Chickens have two legs, horses have four. There is a total of 140 legs. How many horses and how many chickens are there?
The intended solution was to write C for the number of chickens and H for the number of horses. Then we can write down the equations C + H = 50 and 2C + 4 H = 140. From there you can solve the equations using any of the somewhat tedious techniques you find in algebra textbooks.
Even as a kid who generally liked math,
