Let's solve each system of equations step-by-step:
### System 1
[tex]\[
\begin{cases}
y = 3x - 3 \\
y = x^2 + 5x - 2
\end{cases}
\][/tex]
Equate [tex]\( 3x - 3 \)[/tex] to [tex]\( x^2 + 5x - 2 \)[/tex]:
[tex]\[
3x - 3 = x^2 + 5x - 2
\][/tex]
Rearrange to form a standard quadratic equation:
[tex]\[
0 = x^2 + 5x - 2 - 3x + 3 \\
0 = x^2 + 2x + 1
\][/tex]
Solve [tex]\( x^2 + 2x + 1 = 0 \)[/tex]:
[tex]\[
(x + 1)^2 = 0 \implies x = -1
\][/tex]
Substitute [tex]\( x = -1 \)[/tex] back into [tex]\( y = 3x - 3 \)[/tex]:
[tex]\[
y = 3(-1) - 3 = -3 - 3 = -6
\][/tex]
So, the solution is [tex]\((-1, -6)\)[/tex].
### System 2
[tex]\[
\begin{cases}
y = -2x - 7 \\
y = x^2 + 4x - 2
\end{cases}
\][/tex]
Equate [tex]\(-2x - 7\)[/tex] to [tex]\(x^2 + 4x - 2\)[/tex]:
[tex]\[
-2x - 7 = x^2 + 4x - 2
\][/tex]
Rearrange to form a standard quadratic equation:
[tex]\[
0 = x^2 + 4x - 2 + 2x + 7 \\
0 = x^2 + 6x + 5
\][/tex]
Solve [tex]\( x^2 + 6x + 5 = 0 \)[/tex]:
[tex]\[
(x + 5)(x + 1) = 0 \implies x = -5 \text{ or } x = -1
\][/tex]
Substitute [tex]\( x = -5 \)[/tex] into [tex]\( y = -2x - 7 \)[/tex]:
[tex]\[
y = -2(-5) - 7 = 10 - 7 = 3 \implies (-5, 3)
\][/tex]
Substitute [tex]\( x = -1 \)[/tex] into [tex]\( y = -2x - 7 \)[/tex]:
[tex]\[
y = -2(-1) - 7 = 2 - 7 = -5 \implies (-1, -5)
\][/tex]
So, the solutions are [tex]\((-5, 3)\)[/tex] and [tex]\((-1, -5)\)[/tex].
### System 3
[tex]\[
\begin{cases}
y = 2x - 7 \\
y = x^2 + 3x - 1
\end{cases}
\][/tex]
Equate [tex]\( 2x - 7 \)[/tex] to [tex]\( x^2 + 3x - 1 \)[/tex]:
[tex]\[
2x - 7 = x^2 + 3x - 1
\][/tex]
Rearrange to form a standard quadratic equation:
[tex]\[
0 = x^2 + 3x - 1 - 2x + 7 \\
0 = x^2 + x + 6
\][/tex]
Solve [tex]\( x^2 + x + 6 = 0 \)[/tex]:
The discriminant of this quadratic equation is:
[tex]\[
D = b^2 - 4ac = 1^2 - 4(1)(6) = 1 - 24 = -23
\][/tex]
Since the discriminant is negative, there are no real roots. The solutions are complex numbers:
[tex]\[
x = \frac{-1 \pm \sqrt{-23}}{2} = \frac{-1 \pm \sqrt{23}i}{2}
\][/tex]
For [tex]\(x = \frac{-1 - \sqrt{23}i}{2}\)[/tex]:
[tex]\[
y = 2\left(\frac{-1 - \sqrt{23}i}{2}\right) - 7 = -1 - \sqrt{23}i - 7 = -8 - \sqrt{23}i \implies \left( \frac{-1 - \sqrt{23}i}{2}, -8 - \sqrt{23}i \right)
\][/tex]
And for [tex]\(x = \frac{-1 + \sqrt{23}i}{2}\)[/tex]:
[tex]\[
y = 2\left(\frac{-1 + \sqrt{23}i}{2}\right) - 7 = -1 + \sqrt{23}i - 7 = -8 + \sqrt{23}i \implies \left( \frac{-1 + \sqrt{23}i}{2}, -8 + \sqrt{23}i \right)
\][/tex]
The complex solutions are:
[tex]\[
\left( \frac{-1 - \sqrt{23}i}{2}, -8 - \sqrt{23}i \right) \\
\left( \frac{-1 + \sqrt{23}i}{2}, -8 + \sqrt{23}i \right)
\][/tex]
In summary:
### Solutions
1. [tex]\(\left\{\begin{array}{l} y=3 x-3 \\ y=x^2+5 x-2 \end{array}\right.\)[/tex]
[tex]\((-1, -6)\)[/tex]
2. [tex]\(\left\{\begin{array}{l} y=-2 x-7 \\ y=x^2+4 x-2 \end{array}\right.\)[/tex]
[tex]\((-5, 3)\)[/tex], [tex]\((-1, -5)\)[/tex]
3. [tex]\(\left\{\begin{array}{l} y=2 x-7 \\ y=x^2+3 x-1 \end{array}\right.\)[/tex]
[tex]\(\left( \frac{-1 - \sqrt{23}i}{2}, -8 - \sqrt{23}i \right)\)[/tex], [tex]\(\left( \frac{-1 + \sqrt{23}i}{2}, -8 + \sqrt{23}i \right)\)[/tex]
Thus, the final entries for the respective blanks are:
[tex]\[-1, -6, none, none\][/tex]
[tex]\[-5, 3\][/tex], [tex]\[-1, -5\][/tex]
[tex]\(\left( \frac{-1 - \sqrt{23}i}{2}, -8 - \sqrt{23}i \right)\)[/tex], [tex]\(\left( \frac{-1 + \sqrt{23}i}{2}, -8 + \sqrt{23}i \right)\)[/tex]
