An algebra teacher drove by a farmyard full of chickens

Jessie Lee

Jessie Lee

Answered question

2021-12-20

An algebra teacher drove by a farmyard full of chickens and pigs. The teacher happened to notice that there were a total of 100 heads and 270 legs. How many chickens. were there? How many pigs were there?

Answer & Explanation

Jim Hunt

Jim Hunt

Beginner2021-12-21Added 45 answers

Step 1
Given: Total number of heads =100
Total number of legs =270
Step 2
A chiken has one head and two legs.
A pig has one head and four legs.
Let, the number of chikens =x
Number of pigs =y
so, x+y=100
y=100x (i)
and, 2x+4y=270
2x+4(100x)=270
2x+4004x=270
Right2x=270400
Right2x=130
Rightx=1302
Rightx=65
Step 3
put, x=65 in (i)
Righty=10065
Righty=35
so, the number of chikens =65
number of pigs =35
Jenny Sheppard

Jenny Sheppard

Beginner2021-12-22Added 35 answers

So let assume Chicken as C and Pigs as P as per question
Total 100 Heads
so P+C=100
And legs are 270 as Pigs will be having 4 legs and chicken will be having 2 legs (so unfortunate for Chicken Lovers hehe)
4P+2C=270
so Now we have two algebraic equation
P+C=100
4P+2C=270
if P=100C
we can put the value of P in second equation which is
4(100C)+2C=270
4004C+2C=270
4002C=270
2C=400270=130
C=1302=65
Means we have 65 Chickens
now
P+C=100
P+65=100
P=10065
P=35
So Answer is
Pigs are 35
Chickens are 65
Don Sumner

Don Sumner

Skilled2021-12-28Added 184 answers

x=Chickens and y=pigs
2x+4y=270
Since chickens have two legs and pigs have four.
From the first equation we can determine a value for
x+y=100
Subtract y from each side.
x=100y
In the second equation, substitute x with (100y)
2x+4y=270
2(100y)+4y=270
Open the brackets and simplify.
2002y+4y=270
200+2y=270
Subtract 200 from each side.
2y=70
Divide both sides by 2.
y=35, the number of pigs.
In the first equation, substitute y with 35.
x+y=100
x+35=100
Subtract 35 from each side.
x=65, the number of chickens.

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