To solve the system of equations given by:
[tex]\[
\begin{cases}
y = -x^2 + 2 \\
y = x
\end{cases}
\][/tex]
we need to find the point(s) where the two curves intersect. Follow these steps:
1. Set the equations equal to each other:
[tex]\[ y = -x^2 + 2 \][/tex]
[tex]\[ y = x \][/tex]
Since both expressions are equal to [tex]\( y \)[/tex], we can set them equal to each other:
[tex]\[ -x^2 + 2 = x \][/tex]
2. Move everything to one side of the equation:
[tex]\[ -x^2 + 2 - x = 0 \][/tex]
3. Simplify the equation:
Rewriting it to arrange in standard quadratic form [tex]\( ax^2 + bx + c = 0 \)[/tex]:
[tex]\[ -x^2 - x + 2 = 0 \][/tex]
To simplify solving, multiply the entire equation by [tex]\(-1\)[/tex]:
[tex]\[ x^2 + x - 2 = 0 \][/tex]
So, the correct scenario is found in option B: [tex]\( x^2 + x - 2 = 0 \)[/tex].
4. Solve the quadratic equation:
[tex]\[ x^2 + x - 2 = 0 \][/tex]
This can be solved by factoring. We need to find two numbers that multiply to [tex]\(-2\)[/tex] (the constant term) and add up to [tex]\(1\)[/tex] (the coefficient of the [tex]\(x\)[/tex] term). These numbers are [tex]\(2\)[/tex] and [tex]\(-1\)[/tex]:
[tex]\[ (x + 2)(x - 1) = 0 \][/tex]
5. Find the roots:
Set each factor to zero:
[tex]\[ x + 2 = 0 \implies x = -2 \][/tex]
[tex]\[ x - 1 = 0 \implies x = 1 \][/tex]
So the solutions for [tex]\( x \)[/tex] are [tex]\( x = -2 \)[/tex] and [tex]\( x = 1 \)[/tex].
6. Find the corresponding [tex]\( y \)[/tex]-values:
Substitute [tex]\( x = -2 \)[/tex] and [tex]\( x = 1 \)[/tex] into [tex]\( y = x \)[/tex]:
- For [tex]\( x = -2 \)[/tex]:
[tex]\[
y = -2
\][/tex]
- For [tex]\( x = 1 \)[/tex]:
[tex]\[
y = 1
\][/tex]
7. Write the ordered pairs:
The point(s) of intersection (solutions) are the ordered pairs [tex]\((-2, -2)\)[/tex] and [tex]\( (1, 1)\)[/tex].
Answer: The ordered pairs solution for the system of equations is [tex]\((-2, -2)\)[/tex] and [tex]\( (1, 1)\)[/tex].
