Question

Solve the following system.

Answer 1

The solutions are:

$(0, -1)\ and\ (\frac{1}{3}, 0)$

Solution:

Given that,

$y = 3x^2 + 2x-1-------- eqn 1\\\\\3x-y=1 ------- eqn 2$

From eqn 2,

3x - y = 1

y = 3x - 1

Substitute the above in eqn 1

$3x -1 = 3x^2 + 2x - 1\\\\3x^2 + 2x - 3x =0\\\\3x^2 -x = 0\\\\x(3x - 1) = 0\\\\Therefore,\\\\x = 0\\\\And\\\\3x - 1 = 0\\\\3x = 1\\\\x = \frac{1}{3}$

When, x = 0

Substitute x = 0 in eqn 2

3(0) - y = 1

y = -1

When x = 1/3

Substitute x = 1/3 in eqn 2

$3\frac{1}{3} - y = 1\\\\ 1 - y = 1\\\\y = 0$

Thus solutions are:

$(0, -1)\ and\ (\frac{1}{3}, 0)$