To solve the system of equations algebraically, we need to find the values of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] that simultaneously satisfy both the equations:
1. [tex]\( y = -2x - 5 \)[/tex]
2. [tex]\( y = x^2 + x - 9 \)[/tex]
We start by equating the expressions for [tex]\( y \)[/tex] from both equations:
[tex]\[
-2x - 5 = x^2 + x - 9
\][/tex]
Now, we move all the terms to one side of the equation to set it to 0:
[tex]\[
x^2 + x - 9 + 2x + 5 = 0
\][/tex]
Combine like terms:
[tex]\[
x^2 + 3x - 4 = 0
\][/tex]
Next, we solve this quadratic equation using the quadratic formula, which is:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
Here, [tex]\( a = 1 \)[/tex], [tex]\( b = 3 \)[/tex], and [tex]\( c = -4 \)[/tex]. Plug these values into the quadratic formula:
[tex]\[
x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1}
\][/tex]
Simplify under the square root:
[tex]\[
x = \frac{-3 \pm \sqrt{9 + 16}}{2}
\][/tex]
[tex]\[
x = \frac{-3 \pm \sqrt{25}}{2}
\][/tex]
[tex]\[
x = \frac{-3 \pm 5}{2}
\][/tex]
We get two solutions for [tex]\( x \)[/tex]:
1. [tex]\( x = \frac{-3 + 5}{2} = \frac{2}{2} = 1 \)[/tex]
2. [tex]\( x = \frac{-3 - 5}{2} = \frac{-8}{2} = -4 \)[/tex]
Now, we find the corresponding [tex]\( y \)[/tex] values for each [tex]\( x \)[/tex]:
For [tex]\( x = 1 \)[/tex]:
[tex]\[
y = -2(1) - 5 = -2 - 5 = -7
\][/tex]
For [tex]\( x = -4 \)[/tex]:
[tex]\[
y = -2(-4) - 5 = 8 - 5 = 3
\][/tex]
Thus, the solutions to the system of equations are:
[tex]\[
(x, y) = (-4, 3) \quad \text{and} \quad (x, y) = (1, -7)
\][/tex]
Therefore, the final solutions in the given format are:
[tex]\[
\boxed{(-4, 3)},\ \boxed{(1, -7)}
\][/tex]
