Question 439278


let's GOATS be {{{x}}} 
and 
CHICKENS be {{{y}}} 
given:

there are {{{40}}} HEADS....-> {{{40=x+y}}} .........1

{{{130}}} FEET....-> {{{130 =4x+2y}}}................2

{{{40=x+y }}}.........1
{{{130 =4x+2y}}}......2
---------------------solve the system

{{{40=x+y}}} ....-> {{{x=40-y}}}.......substitute in 2

{{{130 =4(40-y)+2y}}}

{{{130 = 160-4y+2y}}}

{{{130-160 =-y}}}

{{{-30 =-2y}}}

{{{2y=30}}}

{{{y=15}}}.............CHICKENS

{{{x=40-15}}}

{{{x=25}}}.............GOATS