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?

Question

Jim Hunt · Accepted Answer

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=100−x (i)  and, ⇒2x+4y=270  ⇒2x+4(100−x)=270  ⇒2x+400−4x=270  Right⇒−2x=270−400  Right⇒−2x=−130  Right⇒x=−130−2  Right⇒x=65  Step 3  put, x=65 in (i)  Right⇒y=100−65  Right⇒y=35  so, the number of chikens =65  number of pigs =35

Jenny Sheppard · Accepted Answer

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=100−C   we can put the value of P in second equation which is   4(100−C)+2C=270   400−4C+2C=270   400−2C=270   2C=400−270=130   C=1302=65   Means we have 65 Chickens   now   P+C=100   P+65=100   P=100−65   P=35   So Answer is   Pigs are 35   Chickens are 65

Don Sumner · Accepted Answer

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=100−y In the second equation, substitute x with (100−y) 2x+4y=270 2(100−y)+4y=270 Open the brackets and simplify. 200−2y+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.

An algebra teacher drove by a farmyard full of chickens

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