>>There are 2 trees in a garden (tree "A" and "B") and on the both trees are some birds...<<
Let x = the number of birds on tree "A".
Let y = the number of birds on tree "B".
>>If half of birds of tree A come to tree B...<<
Then tree A will have only
birds and
tree B will have
birds.
>>it (tree B) will become equal to the birds on tree A...<<
So
which, upon multiplying through by 2,
(1) So in the beginning, tree A has twice as many birds as tree B.
Remember that as statement (1)
>>and if half of birds of tree B comes to tree A<<
Then tree B will have only
birds and
tree A will have
birds.
it (tree A) will become double of the birds on tree B.
So
which, upon multiplying through by 2,
(2) So in the beginning, tree A has one and a half as many birds as tree B.
Remember that as statement (2)
>>so, How many birds are on each tree?<<
There is no solution because statements (1) and (2) are contradictory.
We solve the system:
We substitute 2y for x in 2x=3y
2(2y) = 3y
4y = 3y
y = 0
x = 2y = 2(0) = 0
There are no birds on either tree.
That can't be a solution unless "some birds" can mean "no birds".
Edwin