Question 42465
{{{(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}}}


R^2 at SCC