farmer jane has only pigs and chickens o her farm. she can't remembr how many of each she has but she does know she counted 110 legs and 35 heads. assuming that each animal is normalhow many of each animal does she have
assuming 2 legs per animal, 35 heads would require 70 legs.
Given that there are 110 legs, that means an extra 40 legs, or 20 pairs.
So, there are 20 pigs and 15 chickens
algebraically,
p+c=35
4p+2c=110
p=20
c=15
To find out how many pigs and chickens Farmer Jane has, we need to solve a system of equations based on the given information.
Let's assume that the number of pigs is represented by variable 'p' and the number of chickens is represented by variable 'c'.
Since pigs have 4 legs and chickens have 2 legs, we can set up the equation for the number of legs:
4p + 2c = 110 -- Equation 1
Next, we can set up the equation for the number of heads:
p + c = 35 -- Equation 2
To solve this system of equations, we can use a method called substitution:
Step 1: Solve Equation 2 for one variable (e.g., p):
p = 35 - c
Step 2: Substitute the value of p in Equation 1:
4(35 - c) + 2c = 110
Step 3: Simplify and solve for c:
140 - 4c + 2c = 110
-2c = -30
c = 15
Now that we know that Farmer Jane has 15 chickens, we can substitute this value back into Equation 2 to find the number of pigs:
p + 15 = 35
p = 35 - 15
p = 20
Therefore, Farmer Jane has 20 pigs and 15 chickens on her farm.