A farm raises a total of 220 chickens and pigs. The number of legs of the stock in the farm totals 520. how many chickens and pigs are there on the farm.
using systems of equations to solve
number of chickens --- c
number of pigs ---- p
c+p = 220
2c + 4p = 520 or c + 2p = 260
subtract then
p = 40
back into the first
c+40 = 220
c = 180
180 chickens, 40 pigs
Let's represent the number of chickens as "c" and the number of pigs as "p".
We know that the total number of animals on the farm is 220, so we can write the equation:
c + p = 220 -- Equation 1
We also know that the total number of legs on the farm is 520. Since chickens have 2 legs and pigs have 4 legs, we can write the equation:
2c + 4p = 520 -- Equation 2
To solve the system of equations, we can use the method of substitution or elimination. I'll use the method of substitution.
From Equation 1, we can express c in terms of p:
c = 220 - p
Now substitute this value of c into Equation 2:
2(220 - p) + 4p = 520
440 - 2p + 4p = 520
2p = 520 - 440
2p = 80
p = 80/2
p = 40
Now substitute this value of p back into Equation 1 to find the number of chickens:
c + 40 = 220
c = 220 - 40
c = 180
Therefore, there are 180 chickens and 40 pigs on the farm.
To solve this problem using systems of equations, let's assume the number of chickens as 'x' and the number of pigs as 'y'.
We can set up two equations based on the given information:
Equation 1: x + y = 220 (Total number of chickens and pigs)
Equation 2: 2x + 4y = 520 (Total number of legs)
Now, we have a system of equations. To solve for the values of 'x' and 'y', we can use one of the methods such as substitution or elimination.
Method 1: Substitution
From Equation 1, we can express x in terms of y: x = 220 - y
Substituting this value of x into Equation 2:
2(220 - y) + 4y = 520
440 - 2y + 4y = 520
2y = 520 - 440
2y = 80
y = 40
Now, substituting the value of y back into Equation 1 to solve for x:
x + 40 = 220
x = 220 - 40
x = 180
Therefore, there are 180 chickens and 40 pigs on the farm.