Question 1087782
{{{ xy + 3x^2y + 1 + 2xy^2 + yx^2 + 2 + xy }}}
= {{{ 4x^2y + 2xy^2 + 2xy + 3 }}}
= {{{ highlight( 2xy(2x + y + 1) + 3 ) }}}

I don't see any further simplifications.  
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In response to your 'thank you' message: yes, this expression
3 (x+1) + 4 = 5-3 (x-2)

can be solved for x:
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(3x + 3) + 4 = 5 - (3x - 6)       ( it happens both sides had a 3 that could be distributed)
3x + 7  = 5 - 3x + 6              ( combined 3 + 4 on left,  removed paren's on both sides)
3x + 7  = 11 - 3x                 ( combined 5 + 6 on right)
6x + 7  = 11                      ( added 3x to both sides)
6x  =  4                          ( subtracted 7 from both sides)
  x  = 4/6 = 2/3                  ( divided both sides by 6, then reduced the fraction )
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Check (always a good idea):  
 3(2/3 + 1) + 4 = 2 + 3 + 4 = 9   (left side value)
 5 - 3(2/3 - 2) = 5 - 2 + 6 = 9    (right side value,  matches left side so we're good)

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