To solve the system of equations:
[tex]\[
\begin{cases}
y = 2x + 1 \\
y = x^2 + 2x - 8
\end{cases}
\][/tex]
we can follow these steps:
1. Set the equations equal to each other:
Since both expressions are equal to [tex]\( y \)[/tex], we can set them equal to each other:
[tex]\[
2x + 1 = x^2 + 2x - 8
\][/tex]
2. Simplify and solve for [tex]\( x \)[/tex]:
Subtract [tex]\( 2x \)[/tex] and 1 from both sides to set the equation to zero:
[tex]\[
0 = x^2 - 8 - 1
\][/tex]
Simplify further:
[tex]\[
0 = x^2 - 9
\][/tex]
3. Factor the quadratic equation:
The equation [tex]\( x^2 - 9 = 0 \)[/tex] can be factored as a difference of squares:
[tex]\[
(x - 3)(x + 3) = 0
\][/tex]
4. Solve for [tex]\( x \)[/tex]:
Set each factor equal to zero:
[tex]\[
x - 3 = 0 \quad \text{or} \quad x + 3 = 0
\][/tex]
Solving these gives:
[tex]\[
x = 3 \quad \text{or} \quad x = -3
\][/tex]
5. Substitute [tex]\( x \)[/tex] back into one of the original equations to solve for [tex]\( y \)[/tex]:
We can use the first equation [tex]\( y = 2x + 1 \)[/tex].
For [tex]\( x = 3 \)[/tex]:
[tex]\[
y = 2(3) + 1 = 6 + 1 = 7
\][/tex]
For [tex]\( x = -3 \)[/tex]:
[tex]\[
y = 2(-3) + 1 = -6 + 1 = -5
\][/tex]
6. Write the solutions as ordered pairs:
Thus, the solutions to the system of equations are:
[tex]\[
(3, 7) \quad \text{and} \quad (-3, -5)
\][/tex]
Comparing to the given options:
A. [tex]\((-3,5)\)[/tex] and [tex]\((3,2)\)[/tex]
B. [tex]\((0,1)\)[/tex] and [tex]\((2,5)\)[/tex]
C. [tex]\((-4,0)\)[/tex] and [tex]\((2,0)\)[/tex]
D. [tex]\((-3,-5)\)[/tex] and [tex]\((3,7)\)[/tex]
The correct choice is therefore:
D. [tex]\((-3, -5)\)[/tex] and [tex]\((3, 7)\)[/tex]
