To solve the given system of equations, we must find the values of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] that satisfy both equations simultaneously.
The equations are given as:
1. [tex]\( y = x^2 - 2x - 19 \)[/tex]
2. [tex]\( y + 4x = 5 \)[/tex]
First, we solve the second equation for [tex]\( y \)[/tex]:
[tex]\[ y = 5 - 4x \][/tex]
Now we substitute this expression for [tex]\( y \)[/tex] into the first equation:
[tex]\[ 5 - 4x = x^2 - 2x - 19 \][/tex]
Rearranging the terms to form a quadratic equation:
[tex]\[ x^2 - 2x + 4x - 19 - 5 = 0 \][/tex]
[tex]\[ x^2 + 2x - 24 = 0 \][/tex]
We solve this quadratic equation using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = 2 \)[/tex], and [tex]\( c = -24 \)[/tex]:
[tex]\[ x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-24)}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{-2 \pm \sqrt{4 + 96}}{2} \][/tex]
[tex]\[ x = \frac{-2 \pm \sqrt{100}}{2} \][/tex]
[tex]\[ x = \frac{-2 \pm 10}{2} \][/tex]
This gives us two solutions for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{-2 + 10}{2} = \frac{8}{2} = 4 \][/tex]
[tex]\[ x = \frac{-2 - 10}{2} = \frac{-12}{2} = -6 \][/tex]
Now that we have the values of [tex]\( x \)[/tex], we can find the corresponding [tex]\( y \)[/tex] values by substituting [tex]\( x \)[/tex] back into either of the original equations. Using the second equation [tex]\( y + 4x = 5 \)[/tex]:
For [tex]\( x = 4 \)[/tex]:
[tex]\[ y + 4(4) = 5 \][/tex]
[tex]\[ y + 16 = 5 \][/tex]
[tex]\[ y = 5 - 16 \][/tex]
[tex]\[ y = -11 \][/tex]
For [tex]\( x = -6 \)[/tex]:
[tex]\[ y + 4(-6) = 5 \][/tex]
[tex]\[ y - 24 = 5 \][/tex]
[tex]\[ y = 5 + 24 \][/tex]
[tex]\[ y = 29 \][/tex]
Thus, the solutions to the system of equations are:
[tex]\[ (x, y) = (4, -11) \text{ and } (x, y) = (-6, 29) \][/tex]
The pair of points representing the solution set of this system is [tex]\( (-6, 29) \)[/tex] and [tex]\( (4, -11) \)[/tex].
So, the final answer to be filled in the box is:
[tex]\[ (4, -11) \][/tex]
