The solution above is incorrect due to a careless error:
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 <--This step is wrong because she omitted the
y2 term which vitiated the rest.
Substitute for y in:
Multiply through by 4
This has no rational zeros, but it has 2 real irrational solutions, which
can be approximated by technology. I used a TI-84 graphing calculator.
Then I substituted the approximation values of x in to
find y. The approximations of the solutions are:
(x,y) ≈ (-0.6564052299,-0.2845660870)
(x,y) ≈ (0.40293604482,-0.4188212719)
Edwin