[tex](0, -1)\ and\ (\frac{1}{3}, 0)[/tex]
Solution:
Given that,
[tex]y = 3x^2 + 2x-1-------- eqn 1\\\\\3x-y=1 ------- eqn 2[/tex]
From eqn 2,
3x - y = 1
y = 3x - 1
Substitute the above in eqn 1
[tex]3x -1 = 3x^2 + 2x - 1\\\\3x^2 + 2x - 3x =0\\\\3x^2 -x = 0\\\\x(3x - 1) = 0\\\\Therefore,\\\\x = 0\\\\And\\\\3x - 1 = 0\\\\3x = 1\\\\x = \frac{1}{3}[/tex]
Substitute x = 0 in eqn 2
3(0) - y = 1
y = -1
Substitute x = 1/3 in eqn 2
[tex]3\frac{1}{3} - y = 1\\\\ 1 - y = 1\\\\y = 0[/tex]
Thus solutions are:
[tex](0, -1)\ and\ (\frac{1}{3}, 0)[/tex]