Since there hasn't been a solution, I'll post one for those who are interested.
Let x be the number of chickens which the farmer buys.
Let y be the number of cows which the farmer buys.
Therefore,
20x + 10y = 2000
2x + y = 200
2x= 200-y ... (1)
x= (200-y)/2 = 100 - y/2
Since x...
I noticed that too.
I saw that there hadn't been a solution to the problem and thought that it would be best to give an example of Euler's extended method, in case anyone else was interested.
Let x be the number of bags of corn which the farmer buys.
Let y be the number of bags of rice which the farmer buys.
Here's a solution to the problem.
10x + 20y = 1770 can be simplified to x+2y=177.
Make y the subject:
2y = 177- x ... (1)
y = (177-x)/2 = (88*2+1-x)/2 = 44 + (1-x)/2...