If x2 + y2 + xy = 1 and x + y = 2, then xy =(a) –3(b) 3(c) -32(d) 0

1.	If x2 + y2 + xy = 1 and x + y = 2, then xy =(a) –3(b) 3(c) -32(d) 0
Answer» If x² + y² + xy = 1 and x + y = 2, then xy = (a) –3 (b) 3 (c) $- \frac{3}{2}$ (d) 0

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