To solve the given system of equations:
[tex]\[
\begin{array}{l}
y = 3x^2 + 6x + 4 \\
y = -3x^2 + 4
\end{array}
\][/tex]
we start by setting the two expressions for [tex]\( y \)[/tex] equal to each other:
[tex]\[
3x^2 + 6x + 4 = -3x^2 + 4
\][/tex]
Next, we get all the terms on one side to form a single equation:
[tex]\[
3x^2 + 6x + 4 + 3x^2 - 4 = 0
\][/tex]
Simplify the equation:
[tex]\[
6x^2 + 6x = 0
\][/tex]
Factor out the common term, which is [tex]\( 6x \)[/tex]:
[tex]\[
6x(x + 1) = 0
\][/tex]
This gives us two solutions:
[tex]\[
6x = 0 \quad \text{or} \quad x + 1 = 0
\][/tex]
Solving these, we find:
[tex]\[
x = 0 \quad \text{or} \quad x = -1
\][/tex]
Next, we find the corresponding [tex]\( y \)[/tex]-values for each [tex]\( x \)[/tex]-value by substituting back into one of the original equations. We'll use [tex]\( y = 3x^2 + 6x + 4 \)[/tex].
For [tex]\( x = 0 \)[/tex]:
[tex]\[
y = 3(0)^2 + 6(0) + 4 = 4
\][/tex]
For [tex]\( x = -1 \)[/tex]:
[tex]\[
y = 3(-1)^2 + 6(-1) + 4 = 3(1) - 6 + 4 = 3 - 6 + 4 = 1
\][/tex]
Thus, the solutions to the system of equations are:
[tex]\[
(x, y) = (0, 4) \quad \text{and} \quad (x, y) = (-1, 1)
\][/tex]
So, the final solutions are [tex]\( x = 0 \)[/tex] with [tex]\( y = 4 \)[/tex] and [tex]\( x = -1 \)[/tex] with [tex]\( y = 1 \)[/tex].
