Read the problem below. Then, describe two or more alternate ways to solve this problem.
A farmer sends his daughter and son out into the barnyard to count the number of chickens and pigs. When they return the son says that he counted 200 legs and the daughter says she counted 70 heads. How many pigs and chickens does the farmer have?
A student well versed in algebra might do the following to set up the problem: p = pigs, c = chickens. p + c = 70 (heads) 4p + 2c = 200 (pigs have 4 legs and chickens have 2 legs). These two equations may be used to solve the problem. Students might solve this problem by "guessing and checking," or drawing pictures. Some methods of solving problems might be considered more "efficient." That may be true, but the correct answer can be found using multiple methods. Children think about mathematics in different ways depending on their prior experiences at home and school. By allowing students to think flexibly about numbers, we encourage them to "own" the math forever, instead of "borrowing" until class is over.
1) We can solve this using only one equation:
Let x be the number of chickens. Then, (70-x) is the number of pigs.
Now, pigs have 4 legs and chickens have 2, so the total number of legs is
2x+4(70-x) = 200.
2x + 280 - 4x = ...
Alternative methods to solve a problem are conveyed.