How to solve: There are 40 chickens and cows in the farmyard. There are 100 legs in all. How many are cows and how many are chickens
chickens = 40/2
cows = 100/4
sorry!!!
cows=x
chickens=40-x
100=2(x)+4(40-x)
sorry again!
100=4(x)+2(40-x)
is the right one..
100=4(Γβ)+2(40-Γβ)
50/2 = 25+47=72/4=18+36=54 and there is 25 heads and 2 eyes on each animal πΆ πΆ πΆ πΆ πΆ πΆ πΆ πΆ πΆ πΆ πΆ πΆ πΆ πΆ πΆ πΆ πΆ πΆπΆ πΆ πΆ πΆ πΆ πΆ πΆ πΆ πΆ πΆ πΆ πΆ πΆ πΆ πΆ πΆ πΆ πΆ πΆ πΆ πΆ πΆ πΆ πΆ πΆ πΆ πΆ πΆ πΆ πΆ πΆ
Just divide 50/2 =25+47=72/4=18+36=54 and that is how you get 25 heads and 2 eyes.
To solve this problem, you can approach it using a system of equations. Let's assume the number of chickens is represented by 'C' and the number of cows is represented by 'W'.
Since each chicken has 2 legs and each cow has 4 legs, we can create the following equation based on the given information:
2C + 4W = 100
Now, we need to find a solution where both C and W are integers. To do this, we will need to use some algebraic techniques.
First, let's simplify the equation by dividing it by 2:
C + 2W = 50
Next, we'll isolate one of the variables by using substitution. Let's isolate 'C' by subtracting 2W from both sides of the equation:
C = 50 - 2W
Now, we have an equation solely in terms of 'W'. Since we need both C and W to be integers, we can try different values for 'W' and check if the corresponding 'C' is an integer.
Since the number of animals cannot be negative, let's try different values of 'W' from 0 to 20 and calculate the corresponding value of 'C':
When W = 0, C = 50 - 2(0) = 50
When W = 1, C = 50 - 2(1) = 48
When W = 2, C = 50 - 2(2) = 46
...
By trying different values for 'W', you will find that when W = 10, C = 30, which are both integers. This means that there are 30 chickens and 10 cows on the farmyard.
Therefore, there are 30 chickens and 10 cows.