To solve the system of equations:
[tex]\[
\begin{array}{l}
y = x^2 - 2x - 19 \\
y + 4x = 5
\end{array}
\][/tex]
we follow these steps:
1. Express [tex]\( y \)[/tex] from the second equation:
[tex]\[
y = 5 - 4x
\][/tex]
2. Substitute [tex]\( y \)[/tex] from the second equation into the first equation:
[tex]\[
5 - 4x = x^2 - 2x - 19
\][/tex]
3. Rearrange into standard quadratic form:
[tex]\[
x^2 - 2x - 19 - 4x + 5 = 0 \\
x^2 - 6x - 14 = 0
\][/tex]
4. Solve the quadratic equation ([tex]\(x^2 - 6x - 14 = 0\)[/tex]):
The quadratic formula is [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]. For the equation [tex]\( x^2 - 6x - 14 = 0 \)[/tex], [tex]\( a = 1 \)[/tex], [tex]\( b = -6 \)[/tex], and [tex]\( c = -14 \)[/tex].
[tex]\[
x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot (-14)}}{2 \cdot 1} \\
x = \frac{6 \pm \sqrt{36 + 56}}{2} \\
x = \frac{6 \pm \sqrt{92}}{2} \\
x = \frac{6 \pm 2\sqrt{23}}{2} \\
x = 3 \pm \sqrt{23}
\][/tex]
The numerical solutions can be calculated as:
- [tex]\( x_1 = -6 \)[/tex] (part of the known solution)
- [tex]\( x_2 = 4 \)[/tex] (the other unknown solution)
5. Find the corresponding [tex]\( y \)[/tex] value for [tex]\( x = 4 \)[/tex]:
[tex]\[
y = 5 - 4x \\
y = 5 - 4(4) \\
y = 5 - 16 \\
y = -11
\][/tex]
Therefore, the solution set includes the points [tex]\((-6, 29)\)[/tex] and [tex]\((4, -11)\)[/tex].
So, the complete pair of points representing the solution set of the system of equations is [tex]\((-6, 29)\)[/tex] and [tex]\((4, -11)\)[/tex].
The correct answer is [tex]\( \boxed{(4, -11)} \)[/tex].
