To solve the system of equations algebraically, we need to find the points of intersection of the two equations:
[tex]\[
\left\{\begin{array}{l}
y=-\frac{1}{3} x-6 \\
y=x^2+6x+5
\end{array}\right.
\][/tex]
1. Set the equations equal to each other to determine the values of [tex]\(x\)[/tex], since at the points of intersection the [tex]\(y\)[/tex]-values are the same for both equations:
[tex]\[
-\frac{1}{3} x - 6 = x^2 + 6x + 5
\][/tex]
2. Clear the fraction by multiplying every term by 3:
[tex]\[
- x - 18 = 3x^2 + 18x + 15
\][/tex]
3. Rearrange the equation to set it to zero:
[tex]\[
3x^2 + 18x + 15 + x + 18 = 0
\][/tex]
[tex]\[
3x^2 + 19x + 33 = 0
\][/tex]
4. Solve the quadratic equation [tex]\( 3x^2 + 19x + 33 = 0 \)[/tex]. To solve for [tex]\(x\)[/tex], we use the quadratic formula:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
where [tex]\(a = 3\)[/tex], [tex]\(b = 19\)[/tex], and [tex]\(c = 33\)[/tex].
5. Calculate the discriminant ([tex]\(\Delta\)[/tex]):
[tex]\[
\Delta = b^2 - 4ac = 19^2 - 4 \cdot 3 \cdot 33 = 361 - 396 = -35
\][/tex]
Since the discriminant is negative, there are no real solutions for [tex]\(x\)[/tex].
6. Because there are no real solutions for [tex]\(x\)[/tex], the system has no points of intersection.
There are no real solutions to this system of equations. Therefore, the correct answer is:
( None, None ), ( None, None )
