SOLUTION: Solve each system 2x^2-xy+y^2=8 xy=4

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Question 435357: Solve each system
2x^2-xy+y^2=8
xy=4
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
2x%5E2-xy%2By%5E2=8
xy=4....=>...x=4%2Fy.substitute in 2x%5E2-xy%2By%5E2=8

2%284%2Fy%29%5E2-%284%2Fy%29y%2By%5E2=8....solve for y
2%2816%2Fy%5E2%29-%284y%2Fy%29%2By%5E2=8
2%2816%2Fy%5E2%29-%284cross%28y%29%2Fcross%28y%29%29%2By%5E2=8
2%2816%2Fy%5E2%29-4%2By%5E2=8...both sides divide by 2
2%2816%2Fy%5E2%29%2F2-4%2F2%2By%5E2%2F2=8%2F2
16%2Fy%5E2-+2+%2By%5E2%2F2=4
16%2Fy%5E2%2By%5E2%2F2=4%2B2
16%2Fy%5E2%2By%5E2%2F2=6...both sides divide by 2y%5E2
16%2A2y%5E2%2Fy%5E2%2B+y%5E2%2A2y%5E2%2F2=6%2A2y%5E2
32cross%28y%5E2%29%2Fcross%28y%5E2%29%2B+y%5E2%2Across%282%29y%5E2%2Fcross%282%29=+12y%5E2
32+%2B+y%5E4=+12y%5E2
y%5E4+-12y%5E2%2B32=0
y%5E4+-4y%5E2-8y%5E2%2B32=0
%28y%5E4+-4y%5E2%29-+%288y%5E2-32%29=0
y%5E2%28y%5E2+-4%29-+8%28y%5E2-4%29=0
%28y%5E2-+8%29%28y%5E2-4%29=0
if
y%5E2-+8=0...=>
y%5E2=+8
y=+-sqrt%288%29
y=2.8, or y=+-2.8

now find x

x=4%2F2.8
x=1.43, or x=+-1.43

check:
xy=4
1.43%2A2.8=4
4.004=4
4=4