SOLUTION: (x+y)^2 + (x+3y)^2 - 10(x+3y) - 4(x+y) +29 = 0 , find the value from next expresion 2y^2 + xy

Algebra ->  Expressions -> SOLUTION: (x+y)^2 + (x+3y)^2 - 10(x+3y) - 4(x+y) +29 = 0 , find the value from next expresion 2y^2 + xy      Log On


   

Question 42466:
(x+y)^2 + (x+3y)^2 - 10(x+3y) - 4(x+y) +29 = 0 , find the value from next expresion

2y^2 + xy
Answer by rapaljer(4671) About Me  (Show Source):
You can put this solution on YOUR website!
%28x%2By%29%5E2+%2B+%28x%2B3y%29%5E2+-+10%28x%2B3y%29+-+4%28x%2By%29+%2B29+=+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:


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.

%28x%2By%29%5E2+-+4%28x%2By%29+%2B+4+%2B+%28x%2B3y%29%5E2+-10%28x%2B3y%29+%2B+25+=+0
++%28x%2By-2%29%5E2++%2B+%28x%2B3y-5%29%5E2+=+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%2F2

x+y=2
x%2B%283%2F2%29=2
x=1%2F2

NOW, you wanted 2y^2 + xy=2%2A%289%2F4%29%2B%281%2F2%29%2A%283%2F2%29=+%2818%2F4%29%2B%283%2F4%29=21%2F4

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

%28x%2By%29%5E2+%2B+%28x%2B3y%29%5E2+-+10%28x%2B3y%29+-+4%28x%2By%29+%2B29+=+0
%28x%5E2%2B2xy+%2By%5E2%29+%2B+%28x%5E2%2B6xy%2B9y%5E2%29+-+10x-30y+-+4x-4y+%2B29+=+0
2x%5E2+%2B8xy+%2B+10y%5E2+-14x+-34y+%2B29=+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