Algebra
posted by Lacey
X^2+y^2+6y+3=0
x^2+y^29x6=0
List the solutions:
Please and thank you!

Steve
You have two circles:
x^2 + (y+3)^2 = 6
(x  9/2)^2 + y^2 = 105/4
Naturally, they intersect where
x^2+y^2+6y+3 = x^2+y^29x6
6y = 9x9
y = (3x+3)/2
Subbing that into one of the equations,
x^2+(3x+3)^2/49x6=0
13x^2  18x  15 = 0
x = 1/13 (9±2√69)
y = 3/13 (11±√69)
Hmm. I had expected a bit simpler answer.
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