The system of equations is given by:
[tex]\[
\begin{array}{l}
y = x^2 - 2x - 19 \\
y + 4x = 5
\end{array}
\][/tex]
Step-by-step solution:
1. Substitute the second equation into the first:
Start by solving the second equation for [tex]\( y \)[/tex]:
[tex]\[
y = 5 - 4x
\][/tex]
2. Substitute [tex]\( y \)[/tex] in the first equation:
Replace [tex]\( y \)[/tex] in the first equation:
[tex]\[
5 - 4x = x^2 - 2x - 19
\][/tex]
3. Rearrange the equation:
Bring all terms to one side to form a quadratic equation:
[tex]\[
x^2 - 2x - 19 - 4x + 5 = 0
\][/tex]
Simplify the equation:
[tex]\[
x^2 - 6x - 14 = 0
\][/tex]
4. Solve the quadratic equation:
To solve for [tex]\( x \)[/tex], use the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], here [tex]\( a = 1 \)[/tex], [tex]\( b = -6 \)[/tex], and [tex]\( c = -14 \)[/tex]:
[tex]\[
x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot (-14)}}{2 \cdot 1}
\][/tex]
[tex]\[
x = \frac{6 \pm \sqrt{36 + 56}}{2}
\][/tex]
[tex]\[
x = \frac{6 \pm \sqrt{92}}{2}
\][/tex]
[tex]\[
x = \frac{6 \pm 2\sqrt{23}}{2}
\][/tex]
[tex]\[
x = 3 \pm \sqrt{23}
\][/tex]
Therefore, the [tex]\( x \)[/tex]-values are:
[tex]\[
x = -6 \quad \text{or} \quad x = 4
\][/tex]
5. Find the corresponding [tex]\( y \)[/tex]-values:
Substitute [tex]\( x = -6 \)[/tex] into [tex]\( y = 5 - 4x \)[/tex]:
[tex]\[
y = 5 - 4(-6)
\][/tex]
[tex]\[
y = 5 + 24
\][/tex]
[tex]\[
y = 29
\][/tex]
Substitute [tex]\( x = 4 \)[/tex] into [tex]\( y = 5 - 4x \)[/tex]:
[tex]\[
y = 5 - 4(4)
\][/tex]
[tex]\[
y = 5 - 16
\][/tex]
[tex]\[
y = -11
\][/tex]
Hence, the solution set is [tex]\((-6, 29)\)[/tex] and [tex]\((4, -11)\)[/tex].
So, the correct answer in the box is:
[tex]\[
(4, -11)
\][/tex]
