Answer :
You have animals with 4 legs, and animals with 2 legs. Both animals only have one tail. So, given that they only have one tail, there are only 23 animals in the farmyard.
Now, the equation would be,
[tex] \frac{68}{x} =23[/tex]
[tex]68=23x[/tex]
[tex] \frac{68}{23} =x[/tex]
Now, the equation would be,
[tex] \frac{68}{x} =23[/tex]
[tex]68=23x[/tex]
[tex] \frac{68}{23} =x[/tex]
Answer:
12 chickens and 11 pigs
Step-by-step explanation:
We know that...
Chickens have 2 feet and 1 tail.
Pigs have 4 feet and 1 tail.
Let c represent chickens.
Let p represent pigs.
We need to create two equations. One for feet, and one for tails.
Tails:
Because each animal has 1 tail, we can say that the number of pigs, (p) plus the number of chickens, (c) must equal 23. So...
c + p = 23
Feet:
Because each chicken has two feet, and each pig has 4 feet, we can say that 2 times the number of chickens, (c) plus 4 times the number of pigs, (p) needs to equal 68. So...
2c + 4p = 68
Now we have two equations.
2c + 4p = 68
and
c + p = 23
We can use the substitution method to find c and p. The substitution method involves solving for one of the variables with one of the equations, then placing it into the other equation, so that you only have one variable to solve for, which you can do with simple algebra.
Solve for c
c + p = 23
c = 23 - p
Add that to the other equation.
2c + 4p = 68
2(23 - p) + 4p = 68
Solve for p
46 - 2p + 4p = 68
46 + 2p = 68
2p = 22
p = 11
Now that we know p, we can plug that into one of the equations to find c.
c + p = 23
c + (11) = 23
c = 12
Now we have c and p.
There are 12 chickens and 11 pigs on the farm.
