Question 42466
{{{(x+y)^2 + (x+3y)^2 - 10(x+3y) - 4(x+y) +29 = 0}}}


This looks like a neat problem.  Let's rearrange it and see if this helps sort it out, and subtract 29 from each side while you are at it:
{{{(x+y)^2 - 4(x+y) + _____ + (x+3y)^2 -10(x+3y) + _____ = -29 + _____+_____}}}


Notice that you can "complete the square" (remember "half and square"??) on each of these quantities, by adding 4 on the first blank and 25 on the second blank.
{{{(x+y)^2 - 4(x+y) + 4 + (x+3y)^2 -10(x+3y) + 25 = -29 + 4+25}}}
{{{(x+y)^2 - 4(x+y) + 4 + (x+3y)^2 -10(x+3y) + 25 = 0}}}
{{{  (x+y-2)^2  + (x+3y-5)^2 = 0}}}


Since the sum of two squares equals zero, this means that each quantity squared must equal zero!


x+y-2=0 and x+3y-5=0


x+y=2
x+3y=5


Solve by elimination:
-x-y=-2
x+3y=5


So, 2y = 3, 
{{{y=3/2}}}


x+y=2
{{{x+(3/2)=2}}}
{{{x=1/2}}}


NOW, you wanted 2y^2 + xy={{{2*(9/4)+(1/2)*(3/2)= (18/4)+(3/4)=21/4}}}


It might have been easier to just multiply out the quantities involved and hope for an easy solution to this!


{{{(x+y)^2 + (x+3y)^2 - 10(x+3y) - 4(x+y) +29 = 0}}}
{{{(x^2+2xy +y^2) + (x^2+6xy+9y^2) - 10x-30y - 4x-4y +29 = 0}}}
{{{2x^2 +8xy + 10y^2 -14x -34y +29= 0}}}


Nope!!  That's not going anywhere that I can see!!  Better stick to my first solution.  I don't know how to check it.  You'll have to check it for me!!  I never was any good at math!!!


R^2 at SCC