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Let's solve each system of quadratic equations algebraically by setting the two expressions for $ y $ equal to each other and solving the resulting equation. We'll determine the number of real solutions (by analyzing the discriminant) or find the actual values of $ x $, as required.

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Problem 1:

$$
\begin{cases}
y = -3(x - 1)^2 - 5 \\
y = -2x^2 + 2x - 2
\end{cases}
$$

Step 1: Set the right-hand sides equal:
$$
-3(x - 1)^2 - 5 = -2x^2 + 2x - 2
$$

Step 2: Expand the left side:
$$
-3(x^2 - 2x + 1) - 5 = -3x^2 + 6x - 3 - 5 = -3x^2 + 6x - 8
$$

Now set equal:
$$
-3x^2 + 6x - 8 = -2x^2 + 2x - 2
$$

Step 3: Bring all terms to one side:
$$
(-3x^2 + 6x - 8) - (-2x^2 + 2x - 2) = 0 \\
-3x^2 + 6x - 8 + 2x^2 - 2x + 2 = 0 \\
(-x^2 + 4x - 6) = 0
$$

Multiply by -1:
$$
x^2 - 4x + 6 = 0
$$

Step 4: Discriminant:
$$
D = (-4)^2 - 4(1)(6) = 16 - 24 = -8 < 0
$$

No real solutions.

✔ Answer: 0 real solutions

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Problem 2:

$$
\begin{cases}
y = 2\left(x - \frac{3}{2}\right)^2 - \frac{63}{2} \\
y = -x^2 - 3
\end{cases}
$$

Step 1: Set equal:
$$
2\left(x - \frac{3}{2}\right)^2 - \frac{63}{2} = -x^2 - 3
$$

Step 2: Expand left side:
$$
\left(x - \frac{3}{2}\right)^2 = x^2 - 3x + \frac{9}{4} \\
2(x^2 - 3x + \frac{9}{4}) = 2x^2 - 6x + \frac{18}{4} = 2x^2 - 6x + \frac{9}{2}
$$

Now subtract $ \frac{63}{2} $:
$$
2x^2 - 6x + \frac{9}{2} - \frac{63}{2} = 2x^2 - 6x - \frac{54}{2} = 2x^2 - 6x - 27
$$

Set equal to right side:
$$
2x^2 - 6x - 27 = -x^2 - 3
$$

Bring all to one side:
$$
2x^2 - 6x - 27 + x^2 + 3 = 0 \Rightarrow 3x^2 - 6x - 24 = 0
$$

Divide by 3:
$$
x^2 - 2x - 8 = 0
$$

Step 3: Factor:
$$
(x - 4)(x + 2) = 0 \Rightarrow x = 4, x = -2
$$

Two real solutions.

✔ Answer: 2 real solutions

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Problem 3:

$$
\begin{cases}
y = -5x^2 + 14x + 3 \\
y = -2x^2 + 2x + 3
\end{cases}
$$

Set equal:
$$
-5x^2 + 14x + 3 = -2x^2 + 2x + 3
$$

Subtract right side:
$$
(-5x^2 + 14x + 3) - (-2x^2 + 2x + 3) = 0 \\
-5x^2 + 14x + 3 + 2x^2 - 2x - 3 = 0 \\
-3x^2 + 12x = 0
$$

Factor:
$$
-3x(x - 4) = 0 \Rightarrow x = 0, x = 4
$$

Two real solutions.

✔ Answer: 2 real solutions

---

Problem 4:

$$
\begin{cases}
y = -8\left(x - \frac{1}{4}\right)^2 - \frac{9}{2} \\
y = -4x^2 + 4x - 4
\end{cases}
$$

Step 1: Expand left side:
$$
\left(x - \frac{1}{4}\right)^2 = x^2 - \frac{1}{2}x + \frac{1}{16} \\
-8(x^2 - \frac{1}{2}x + \frac{1}{16}) = -8x^2 + 4x - \frac{8}{16} = -8x^2 + 4x - \frac{1}{2}
$$

Now subtract $ \frac{9}{2} $:
$$
-8x^2 + 4x - \frac{1}{2} - \frac{9}{2} = -8x^2 + 4x - 5
$$

Set equal to right side:
$$
-8x^2 + 4x - 5 = -4x^2 + 4x - 4
$$

Subtract right side:
$$
(-8x^2 + 4x - 5) - (-4x^2 + 4x - 4) = -4x^2 - 1 = 0
\Rightarrow -4x^2 = 1 \Rightarrow x^2 = -\frac{1}{4}
$$

No real solution.

✔ Answer: DNE

---

Problem 5:

$$
\begin{cases}
y = 7x^2 + 18x + 13 \\
y = 3x^2 + 2x - 3
\end{cases}
$$

Set equal:
$$
7x^2 + 18x + 13 = 3x^2 + 2x - 3
$$

Subtract:
$$
(7x^2 - 3x^2) + (18x - 2x) + (13 + 3) = 0 \\
4x^2 + 16x + 16 = 0
$$

Divide by 4:
$$
x^2 + 4x + 4 = 0 \Rightarrow (x + 2)^2 = 0 \Rightarrow x = -2
$$

One real solution (double root).

✔ Answer: $ x = -2 $

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Problem 6:

$$
\begin{cases}
y = 6x^2 - x + 2 \\
y = 2x^2 - 3x - 1
\end{cases}
$$

Set equal:
$$
6x^2 - x + 2 = 2x^2 - 3x - 1
$$

Subtract:
$$
(6x^2 - 2x^2) + (-x + 3x) + (2 + 1) = 0 \\
4x^2 + 2x + 3 = 0
$$

Discriminant:
$$
D = (2)^2 - 4(4)(3) = 4 - 48 = -44 < 0
$$

No real solutions.

✔ Answer: DNE

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✔ Final Answers:

1. 0 real solutions
2. 2 real solutions
3. 2 real solutions
4. DNE
5. $ x = -2 $
6. DNE

Let me know if you'd like the corresponding $ y $-values for problems 5 and 6!