Explanation:
I believe there is a typo in the second equation. It should only be $y=9x-4$.
To solve the system of equations, we can substitute the first equation into the second equation for $y$. This gives us:
\begin{align*}
y &= 9x-4 \\
3x^2+2 &= 9x-4 \\
3x^2-9x+6 &= 0 \\
x^2-3x+2 &= 0 \\
(x-1)(x-2) &= 0
\end{align*}
So, $x=1$ or $x=2$.
To find the corresponding values of $y$, we can substitute these values of $x$ into the first equation:
\begin{align*}
\text{If } x=1: \quad y &= 3(1)^2+2 = 5\\
\text{If } x=2: \quad y &= 3(2)^2+2 = 14
\end{align*}
Therefore, the solutions to the system of equations are $(1,5)$ and $(2,14)$.