Answer :
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]
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]
