Question

puntos de intersección de x²+y²-4y-14=0 ;x+y-8=0

Answer 1

x^2+y^2-4y-14 = 0 ; x + y - 8 = 0
.................................; x = 8 - y
(8 - y)^2 + y^2 - 4y - 14 = 0
64 - 16y + y^2 + y^2 - 4y-14 = 0
2y^2-20y+50 = 0
y^2 - 10y + 25 = 0
(y-5)(y-5) = 0

y = 5; x = 8 - 5
.........; x = 3