Question 1178993
3𝑥^2 − 2𝑥𝑦 + 𝑦^2 − 1 = 0 and 

𝑥 =(2𝑦+1)/𝑥
x^2 = (2y+1)

3𝑥^2 − 2𝑥𝑦 + 𝑦^2 − 1 = 0 
3(2y+1)^2 -2y(2y+1) -1=0
3(4y^2+4y+1) -4y^2-2y -1=0
12y^2+12y +3 -4y^2-2y-1=0
8y^2+10y+2=0
8y^2+8y+2y+2=0
8y(y+1) +2(y+1)=0
(8y+2)(y+1)=0
y=-1/4 or y=-1
Plug the values of y in x^2=(2y+1)
and get values of x for each y.