Let's solve the given systems of equations algebraically.
First System:
[tex]\[
\left\{\begin{array}{l}
y = 3x - 3 \\
y = x^2 + 5x - 2
\end{array}\right.
\][/tex]
To find the points of intersection, set the equations equal to each other:
[tex]\[
3x - 3 = x^2 + 5x - 2
\][/tex]
Rearrange the equation to set it to zero:
[tex]\[
x^2 + 5x - 2 - 3x + 3 = 0 \\
x^2 + 2x + 1 = 0
\][/tex]
Factor the quadratic equation:
[tex]\[
(x + 1)^2 = 0
\][/tex]
This gives:
[tex]\[
x = -1
\][/tex]
Plug [tex]\(x = -1\)[/tex] back into the first equation [tex]\(y = 3x - 3\)[/tex]:
[tex]\[
y = 3(-1) - 3 = -6
\][/tex]
So, the solution for the first system is:
[tex]\[
(-1, -6)
\][/tex]
Second System:
[tex]\[
\left\{\begin{array}{l}
y = -2x - 7 \\
y = x^2 + 4x - 2
\end{array}\right.
\][/tex]
Set the equations equal to each other:
[tex]\[
-2x - 7 = x^2 + 4x - 2
\][/tex]
Rearrange the equation to set it to zero:
[tex]\[
x^2 + 4x - 2 + 2x + 7 = 0 \\
x^2 + 6x + 5 = 0
\][/tex]
Factor the quadratic equation:
[tex]\[
(x + 1)(x + 5) = 0
\][/tex]
This gives:
[tex]\[
x = -1 \quad \text{or} \quad x = -5
\][/tex]
For [tex]\(x = -1\)[/tex]:
[tex]\[
y = -2(-1) - 7 = -5
\][/tex]
For [tex]\(x = -5\)[/tex]:
[tex]\[
y = -2(-5) - 7 = 3
\][/tex]
So, the solutions for the second system are:
[tex]\[
(-5, 3) \quad \text{and} \quad (-1, -5)
\][/tex]
Third System:
[tex]\[
\left\{\begin{array}{l}
y = 2x - 7 \\
y = x^2 + 3x - 1
\end{array}\right.
\][/tex]
Set the equations equal to each other:
[tex]\[
2x - 7 = x^2 + 3x - 1
\][/tex]
Rearrange the equation to set it to zero:
[tex]\[
x^2 + 3x - 1 - 2x + 7 = 0 \\
x^2 + x + 6 = 0
\][/tex]
This quadratic equation has no real solutions, as its discriminant is negative:
[tex]\[
\Delta = 1^2 - 4(1)(6) = 1 - 24 = -23
\][/tex]
The solutions to this quadratic are complex (non-real):
[tex]\[
x = \frac{-1 \pm \sqrt{-23}}{2} = \frac{-1 \pm \sqrt{23}i}{2}
\][/tex]
Thus, the points of intersection (in terms of complex numbers) are:
[tex]\[
\left(-\frac{1}{2} - \frac{\sqrt{23}i}{2}, -8 - \sqrt{23}i\right) \quad \text{and} \quad \left(-\frac{1}{2} + \frac{\sqrt{23}i}{2}, -8 + \sqrt{23}i\right)
\][/tex]
Therefore, summarizing the solutions for each system:
[tex]\[
\left\{\begin{array}{l}
y = 3x - 3 \\
y = x^2 + 5x - 2
\end{array}\right.
\Rightarrow (-1, -6)
\][/tex]
[tex]\[
\left\{\begin{array}{l}
y = -2x - 7 \\
y = x^2 + 4x - 2
\end{array}\right.
\Rightarrow (-5, 3) \; \text{and} \; (-1, -5)
\][/tex]
[tex]\[
\left\{\begin{array}{l}
y = 2x - 7 \\
y = x^2 + 3x - 1
\end{array}\right.
\Rightarrow \left(-\frac{1}{2} - \frac{\sqrt{23}i}{2}, -8 - \sqrt{23}i\right) \; \text{and} \; \left(-\frac{1}{2} + \frac{\sqrt{23}i}{2}, -8 + \sqrt{23}i\right)
\][/tex]
Given the required forms:
[tex]\[
( (-1, -6), (-5, 3), (-1, -5), \left(-\frac{1}{2} - \frac{\sqrt{23}i}{2}, -8 - \sqrt{23}i\right), \left(-\frac{1}{2} + \frac{\sqrt{23}i}{2}, -8 + \sqrt{23}i\right) )
\][/tex]
