Question:
If \[ x^2+y^2+xy=1 \] and \[ x+y=2, \] then \[ xy= \]
(a) \(-3\)
(b) \(3\)
(c) \(-\frac{3}{2}\)
(d) \(0\)
Solution:
\[ (x+y)^2=x^2+y^2+2xy \]
\[ 2^2=x^2+y^2+2xy \]
\[ 4=(x^2+y^2+xy)+xy \]
\[ 4=1+xy \]
\[ xy=3 \]
\[ \boxed{3} \]
If \[ x^2+y^2+xy=1 \] and \[ x+y=2, \] then \[ xy= \]
(a) \(-3\)
(b) \(3\)
(c) \(-\frac{3}{2}\)
(d) \(0\)
\[ (x+y)^2=x^2+y^2+2xy \]
\[ 2^2=x^2+y^2+2xy \]
\[ 4=(x^2+y^2+xy)+xy \]
\[ 4=1+xy \]
\[ xy=3 \]
\[ \boxed{3} \]