Afarmer raises cows and chickens. there are a total of 114 legs and 42 animals 4y = cow legs and 2x = chicken legs. how many cows and chickens the farmer has

X+y=42, 4y+2x=114

y=42-x

4(42-x)+2x=114
168-4x+2x=114
168-2x=114
168-114=2x
54=2x
54/2=x
27=x

plug 27 into y=42-x

y=42-27
y=15

Use algebraic formula:
(Animal Type 1:X)x(Number of Legs on that type) + (Animal Type 2:Y)x(Number of Legs on that type) = Total number of legs

2x+4y = 114
Also: X+Y = 42
Divide he equation 2x+4y=114 by 2 on both sides = x+2y=57
If X+Y = 42, and x+2y=57 then Y must = 15 (subtract X+Y from one side and 42 from the other)
15 cows 27 chickens = 114 legs 42 heads

This formula works regardless of how many ‘legs’ the creature type has. Just plug in the appropriate value in the original equation to match the new criteria.

10 will be the number of adult tickets and 11 will be the number of children tickets. i hope i could ! good luck : )