Question 1144295
>>There are 2 trees in a garden (tree "A" and "B") and on the both trees are some birds...<<
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Let x = the number of birds on tree "A".
Let y = the number of birds on tree "B".
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>>If half of birds of tree A come to tree B...<<
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Then tree A will have only {{{expr(1/2)x}}} birds and 
tree B will have {{{y+expr(1/2)x}}} birds.
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>>it (tree B) will become equal to the birds on tree A...<< 
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So {{{y+expr(1/2)x=x}}} which, upon multiplying through by 2,

{{{2y+x=2x}}}
{{{2y=x}}}

(1)  So in the beginning, tree A has twice as many birds as tree B.

Remember that as statement (1)
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>>and if half of birds of tree B comes to tree A<< 
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Then tree B will have only {{{expr(1/2)y}}} birds and 
tree A will have {{{x+expr(1/2)y}}} birds.
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it (tree A) will become double of the birds on tree B.
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So {{{x+expr(1/2)y=2y}}} which, upon multiplying through by 2,

{{{2x+y=4y}}}
{{{2x=3y}}}
{{{x=1.5y}}}
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(2)  So in the beginning, tree A has one and a half as many birds as tree B.

Remember that as statement (2)
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>>so, How many birds are on each tree?<<
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There is no solution because statements (1) and (2) are contradictory.

We solve the system:

{{{system(2y=x,2x=3y)}}}

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</pre>