How many alligators did each breeder have?

Question

Two alligator breeders were talking one day, and in the course of their
conversation they noticed a curious coincidence. If alligator breeder #1
were to sell alligator breeder #2 seven alligators, then breeder #1 would
have exactly as many alligators as breeder #2. On the other hand, if
alligator breeder #2 were to sell alligator breeder #1 seven alligators,
then breeder #1 would have exactly twice as many alligators as breeder #2.
(Please show work)

Answer 1

Take each statement and make an equation:

1.) If alligator breeder #1 were to sell alligator breeder #2 seven alligators, then breeder #1 would have exactly as many alligators as breeder #2.

b1 - 7 = b2 + 7 (b1 loses 7 gators, b2 gains 7)

2.) If alligator breeder #2 were to sell alligator breeder #1 seven alligators, then breeder #1 would have exactly twice as many alligators as breeder #2.

b1 + 7 = 2(b2 - 7) (b1 gains 7 gators, b2 loses 7, and b1 now has twice as many as b2).

3.) Isolate either b1 or b2. Use eqn. 1.
b1 - 7 = b2 + 7
b1 - 7 + 7 = b2 + 7 + 7
b1 = b2 + 14

4.) Substitute the value for b1 into eqn 2.
b1 + 7 = 2(b2 - 7)
b2 + 14 + 7 = 2(b2 - 7)
b2 - b2 + 21 - 21 = 2(b2 - 7) - 21 - b2
0 = 2b2 - 14 -21 - b2
35 = b2 - 35 + 35
35 = b2 (solution)

b1 - 7 = b2 + 7
b1 - 7 = 35 + 7
b1 = 35 + 7 + 7 = 49 (solution)

Answer 2

let no. of alligators breeder #1 has = x
let no. of alligators breeder #2 has = y

situation (1)
x - 7 = y + 7
x = y + 14 ------(a)

situation (2)
2(y - 7) = x + 7
2y - 14 = x + 7
x = 2y - 21 --------(b)

equating (a) and (b),
y + 14 = 2y - 21
y = 35

using (a)
x = y + 14 = 35 + 14 = 49

therefore
let no. of alligators breeder #1 has = x = 49
let no. of alligators breeder #2 has = y = 35

Answer 3

Breeder1=x=21
breeder2=y=14

1. x=y+7
2. x+7=2y
from 2 x=2y-7
insert this in 1.
we get 2y-7=y+7 ---> 2y-y=7+7 --->y=14
and from1 x=14+7=21