Let's solve the following system of equations algebraically:
[tex]$
\left\{\begin{array}{l}
y=\frac{1}{3} x-3 \\
y=x^2+5 x+7
\end{array}\right.
$[/tex]
First, we set the right-hand sides of the two equations equal to each other since they both equal [tex]\( y \)[/tex]:
[tex]\[
\frac{1}{3} x - 3 = x^2 + 5x + 7
\][/tex]
Next, we'll bring all terms to one side to set the equation to 0:
[tex]\[
0 = x^2 + 5x + 7 - \frac{1}{3} x + 3
\][/tex]
Combine like terms:
[tex]\[
0 = x^2 + \left(5 - \frac{1}{3}\right)x + 10
\][/tex]
Simplify the coefficients:
[tex]\[
0 = x^2 + \frac{15}{3}x - \frac{1}{3}x + 10 \\
0 = x^2 + \frac{14}{3}x + 10
\][/tex]
Multiply through by 3 to clear the fractions:
[tex]\[
0 = 3x^2 + 14x + 30
\][/tex]
Now, we solve the quadratic equation using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]. Here, [tex]\( a = 3 \)[/tex], [tex]\( b = 14 \)[/tex], and [tex]\( c = 30 \)[/tex]:
[tex]\[
x = \frac{-14 \pm \sqrt{14^2 - 4 \cdot 3 \cdot 30}}{2 \cdot 3} \\
x = \frac{-14 \pm \sqrt{196 - 360}}{6} \\
x = \frac{-14 \pm \sqrt{-164}}{6} \\
x = \frac{-14 \pm \sqrt{164}i}{6}
\][/tex]
Simplify the expression under the square root:
[tex]\[
x = \frac{-14 \pm 2\sqrt{41}i}{6} \\
x = \frac{-7 \pm \sqrt{41}i}{3}
\][/tex]
Thus, we have two solutions for [tex]\( x \)[/tex]:
[tex]\[
x = \frac{-7}{3} - \frac{\sqrt{41}i}{3} \\
x = \frac{-7}{3} + \frac{\sqrt{41}i}{3}
\][/tex]
Next, we find the corresponding [tex]\( y \)[/tex]-values by substituting each [tex]\( x \)[/tex] back into the first equation [tex]\( y = \frac{1}{3} x - 3 \)[/tex]:
For [tex]\( x = \frac{-7}{3} - \frac{\sqrt{41}i}{3} \)[/tex]:
[tex]\[
y = \frac{1}{3} \left(\frac{-7}{3} - \frac{\sqrt{41}i}{3}\right) - 3 \\
y = \frac{-7}{9} - \frac{\sqrt{41}i}{9} - 3 \\
y = - \frac{7}{9} - \frac{\sqrt{41}i}{9} - \frac{27}{9} \\
y = \frac{-34}{9} - \frac{\sqrt{41}i}{9}
\][/tex]
And for [tex]\( x = \frac{-7}{3} + \frac{\sqrt{41}i}{3} \)[/tex]:
[tex]\[
y = \frac{1}{3} \left(\frac{-7}{3} + \frac{\sqrt{41}i}{3}\right) - 3 \\
y = \frac{-7}{9} + \frac{\sqrt{41}i}{9} - 3 \\
y = - \frac{7}{9} + \frac{\sqrt{41}i}{9} - \frac{27}{9} \\
y = \frac{-34}{9} + \frac{\sqrt{41}i}{9}
\][/tex]
Therefore, the solutions to the system of equations are:
[tex]\[
\left(\frac{-7}{3} - \frac{\sqrt{41}i}{3}, \frac{-34}{9} - \frac{\sqrt{41}i}{9}\right) \text{ and }\left(\frac{-7}{3} + \frac{\sqrt{41}i}{3}, \frac{-34}{9} + \frac{\sqrt{41}i}{9}\right)
\][/tex]
Since these solutions are complex, we do not have any real solutions for the given system of equations. Thus, the answers are:
[tex]\[ \text{(none, none) (none, none)} \][/tex]
