To solve the system of equations given by:
[tex]\[
\begin{array}{l}
y = 2x^2 + 6x + 4 \\
y = -4x^2 + 4
\end{array}
\][/tex]
we need to find the points [tex]\((x, y)\)[/tex] where both equations intersect. This can be done by setting the two equations equal to each other. Here’s the step-by-step procedure:
1. Set the equations equal to each other:
[tex]\[
2x^2 + 6x + 4 = -4x^2 + 4
\][/tex]
2. Move all terms to one side to set the equation to zero:
[tex]\[
2x^2 + 6x + 4 + 4x^2 - 4 = 0
\][/tex]
Combine like terms:
[tex]\[
6x^2 + 6x = 0
\][/tex]
3. Factor the equation:
First, factor out the greatest common factor, which is [tex]\(6x\)[/tex]:
[tex]\[
6x(x + 1) = 0
\][/tex]
4. Solve for [tex]\(x\)[/tex]:
Set each factor equal to zero:
[tex]\[
6x = 0 \quad \text{or} \quad x + 1 = 0
\][/tex]
For [tex]\(6x = 0\)[/tex]:
[tex]\[
x = 0
\][/tex]
For [tex]\(x + 1 = 0\)[/tex]:
[tex]\[
x = -1
\][/tex]
5. Find the corresponding [tex]\(y\)[/tex]-values for each [tex]\(x\)[/tex]-value:
Substitute [tex]\(x = 0\)[/tex] into either original equation (we’ll use the first one):
[tex]\[
y = 2(0)^2 + 6(0) + 4 = 4
\][/tex]
So, the point [tex]\((0, 4)\)[/tex] is a solution.
Substitute [tex]\(x = -1\)[/tex] into either original equation (we’ll use the first one again):
[tex]\[
y = 2(-1)^2 + 6(-1) + 4
\][/tex]
[tex]\[
y = 2(1) - 6 + 4
\][/tex]
[tex]\[
y = 2 - 6 + 4
\][/tex]
[tex]\[
y = 0
\][/tex]
So, the point [tex]\((-1, 0)\)[/tex] is a solution.
Summary:
The solutions to the system of equations are the points where the two equations intersect, which are:
[tex]\[
(x, y) = (-1, 0) \quad \text{and} \quad (x, y) = (0, 4)
\][/tex]
