To solve the system of equations:
[tex]\[
\begin{cases}
y = 2x - 3 \\
y = x^2 - 2x - 8
\end{cases}
\][/tex]
we need to find the points [tex]\((x, y)\)[/tex] where the two curves intersect.
### Step 1: Set the equations equal to each other
Since both equations are equal to [tex]\(y\)[/tex], we set them equal to each other to find the [tex]\(x\)[/tex] values at the points of intersection:
[tex]\[
2x - 3 = x^2 - 2x - 8
\][/tex]
### Step 2: Rearrange the equation to standard quadratic form
Move all terms to one side to set the equation to zero:
[tex]\[
0 = x^2 - 2x - 2x - 8 + 3
\][/tex]
Combine like terms:
[tex]\[
x^2 - 4x - 5 = 0
\][/tex]
### Step 3: Solve the quadratic equation
This is a quadratic equation of the form [tex]\(ax^2 + bx + c = 0\)[/tex]. We can solve it using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 1\)[/tex], [tex]\(b = -4\)[/tex], and [tex]\(c = -5\)[/tex].
Calculate the discriminant:
[tex]\[
\Delta = b^2 - 4ac = (-4)^2 - 4(1)(-5) = 16 + 20 = 36
\][/tex]
Since the discriminant is positive, there are two real solutions:
[tex]\[
x = \frac{-(-4) \pm \sqrt{36}}{2(1)} = \frac{4 \pm 6}{2}
\][/tex]
Calculate the two solutions:
[tex]\[
x = \frac{4 + 6}{2} = \frac{10}{2} = 5
\][/tex]
[tex]\[
x = \frac{4 - 6}{2} = \frac{-2}{2} = -1
\][/tex]
### Step 4: Find the corresponding [tex]\(y\)[/tex] values for each [tex]\(x\)[/tex]
Substitute [tex]\(x = 5\)[/tex] back into one of the original equations, say [tex]\(y = 2x - 3\)[/tex]:
[tex]\[
y = 2(5) - 3 = 10 - 3 = 7
\][/tex]
So one point of intersection is [tex]\((5, 7)\)[/tex].
Next, substitute [tex]\(x = -1\)[/tex] back into the same original equation:
[tex]\[
y = 2(-1) - 3 = -2 - 3 = -5
\][/tex]
So the other point of intersection is [tex]\((-1, -5)\)[/tex].
### Conclusion
The two points of intersection are [tex]\((-1, -5)\)[/tex] and [tex]\((5, 7)\)[/tex].
Therefore, the answer is:
A. [tex]\((-1, -5)\)[/tex] and [tex]\((5, 7)\)[/tex]
