To solve the system of equations given by
[tex]\[
\begin{cases}
y = 2x + 4 \\
y = x^2 + x - 2
\end{cases}
\][/tex]
we need to find the points [tex]\((x, y)\)[/tex] that satisfy both equations simultaneously.
Step 1: Set the equations equal to each other
Since both equations equal [tex]\(y\)[/tex], we can set them equal to each other:
[tex]\[
2x + 4 = x^2 + x - 2
\][/tex]
Step 2: Rearrange into standard quadratic form
To solve for [tex]\(x\)[/tex], we first rearrange all terms to one side of the equation:
[tex]\[
0 = x^2 + x - 2 - 2x - 4
\][/tex]
Simplify:
[tex]\[
0 = x^2 - x - 6
\][/tex]
Step 3: Factor the quadratic equation
Next, we factor the quadratic equation:
[tex]\[
x^2 - x - 6 = (x - 3)(x + 2) = 0
\][/tex]
Step 4: Solve for [tex]\(x\)[/tex]
Set each factor equal to zero and solve for [tex]\(x\)[/tex]:
[tex]\[
x - 3 = 0 \quad \Rightarrow \quad x = 3
\][/tex]
[tex]\[
x + 2 = 0 \quad \Rightarrow \quad x = -2
\][/tex]
Step 5: Substitute [tex]\(x\)[/tex] values into one of the original equations to find [tex]\(y\)[/tex]
Now, we substitute [tex]\(x = 3\)[/tex] and [tex]\(x = -2\)[/tex] into the first equation [tex]\(y = 2x + 4\)[/tex]:
For [tex]\(x = 3\)[/tex]:
[tex]\[
y = 2(3) + 4 = 6 + 4 = 10
\][/tex]
So, one point is [tex]\((3, 10)\)[/tex].
For [tex]\(x = -2\)[/tex]:
[tex]\[
y = 2(-2) + 4 = -4 + 4 = 0
\][/tex]
So, another point is [tex]\((-2, 0)\)[/tex].
Conclusion
The solutions to the system of equations are [tex]\((3, 10)\)[/tex] and [tex]\((-2, 0)\)[/tex]. We check these solutions against the given choices:
A. [tex]\((-2,0)\)[/tex] and [tex]\((3,10)\)[/tex]
B. [tex]\((2,0)\)[/tex] and [tex]\((-3,-10)\)[/tex]
C. [tex]\((-3,-10)\)[/tex] and [tex]\((-2,0)\)[/tex]
D. [tex]\((-3,10)\)[/tex] and [tex]\((2,0)\)[/tex]
Choice A matches our solutions exactly.
Answer: A. [tex]\((-2,0)\)[/tex] and [tex]\((3,10)\)[/tex]
