Hello, Kalayh, Q; How many pigs and how many chickens are there? This problem can be worked as an Algebra system of equations: Let x=# of cows y=# of chickens Total pigs and chickens=40 total legs=100 Ask yourself: How many legs does a cow have? How many legs does a chicken have? Then 4x+2y=100 x +y= 40 Multiply bottom row by -2 in order to eliminate one variable. 4x + 2y=100 -2x + -2y=-80 Add equations together. 2x=20 x=20/2 =10 Now, just substitute x=10 into the easier equation up there, and you can find y. I will let you finish the problem. It is also nice to check your answers by substituting the values of "x" and "y" back into the original equations in the system. If one side of the equation equals the other side of the equal sign, then you know that your answers are correct. However, you would need to substitute the values into both equations to check them. Susan C.