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(×)+2(40-×)

50/2 = 25+47=72/4=18+36=54 and there is 25 heads and 2 eyes on each animal 😶 😶 😶 😶 😶 😶 😶 😶 😶 😶 😶 😶 😶 😶 😶 😶 😶 😶😶 😶 😶 😶 😶 😶 😶 😶 😶 😶 😶 😶 😶 😶 😶 😶 😶 😶 😶 😶 😶 😶 😶 😶 😶 😶 😶 😶 😶 😶 😶

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.