Question

Puntos de interseccion de la recta x+y-1=0 con la parabola x2-4xy+1 , .

Answer 1

x=1-y
(1-y)+4(1-y)y+1=0
1-y+4y-4y^2+1=0
4y^2-3y-2=0
y=(3+(41)^(1/2))/8    x=(5-(41)^(1/2))/8
y:(3-(41)^(1/2))/8      x=(5-(41)^(1/2))/8

Answer 2

x=1-y

(1-y)+4(1-y)y+1=0

1-y+4y-4y^2+1=0

4y^2-3y-2=0

y=(3+(41)^(1/2))/8

y:(3-(41)^(1/2))/8

x=(5-(41)^(1/2))/8

x=(5-(41)^(1/2))/8