Solve the given algebraic equations involving fractions with variables in the denominators.

Let's solve each of the four rational equations step by step.

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1.

$$
\frac{-3}{x+1} = \frac{4}{x-1}
$$

Step 1: Cross-multiply to eliminate denominators:
$$
-3(x - 1) = 4(x + 1)
$$

Step 2: Expand both sides:
$$
-3x + 3 = 4x + 4
$$

Step 3: Move all terms to one side:
$$
3 - 4 = 4x + 3x \Rightarrow -1 = 7x
$$

Step 4: Solve for $ x $:
$$
x = -\frac{1}{7}
$$

Check for restrictions:
Denominators: $ x + 1 \neq 0 \Rightarrow x \neq -1 $, and $ x - 1 \neq 0 \Rightarrow x \neq 1 $.
$ x = -\frac{1}{7} $ is allowed.

✔ Solution: $ \boxed{x = -\frac{1}{7}} $

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2.

$$
\frac{x}{x^2 - 8} = \frac{2}{x}
$$

Step 1: Note that $ x^2 - 8 = (x - \sqrt{8})(x + \sqrt{8}) $, but we can proceed algebraically.

Cross-multiply:
$$
x \cdot x = 2(x^2 - 8)
$$

$$
x^2 = 2x^2 - 16
$$

Step 2: Bring all terms to one side:
$$
x^2 - 2x^2 = -16 \Rightarrow -x^2 = -16 \Rightarrow x^2 = 16
$$

Step 3: Solve:
$$
x = \pm 4
$$

Check restrictions:
Denominators: $ x^2 - 8 \neq 0 \Rightarrow x \neq \pm \sqrt{8} \approx \pm 2.828 $, and $ x \neq 0 $.
So $ x = 4 $ and $ x = -4 $ are both acceptable.

✔ Solutions: $ \boxed{x = 4} $ or $ \boxed{x = -4} $

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3.

$$
\frac{x}{2x + 7} = \frac{x - 5}{x - 1}
$$

Step 1: Cross-multiply:
$$
x(x - 1) = (x - 5)(2x + 7)
$$

Step 2: Expand both sides:

Left: $ x^2 - x $

Right: $ (x)(2x + 7) - 5(2x + 7) = 2x^2 + 7x - 10x - 35 = 2x^2 - 3x - 35 $

So:
$$
x^2 - x = 2x^2 - 3x - 35
$$

Step 3: Bring all terms to one side:
$$
x^2 - x - 2x^2 + 3x + 35 = 0 \Rightarrow -x^2 + 2x + 35 = 0
$$

Multiply by -1:
$$
x^2 - 2x - 35 = 0
$$

Step 4: Factor:
$$
(x - 7)(x + 5) = 0
\Rightarrow x = 7 \quad \text{or} \quad x = -5
$$

Check restrictions:
Denominators:
- $ 2x + 7 \neq 0 \Rightarrow x \neq -\frac{7}{2} $
- $ x - 1 \neq 0 \Rightarrow x \neq 1 $

Both $ x = 7 $ and $ x = -5 $ are valid.

✔ Solutions: $ \boxed{x = 7} $ or $ \boxed{x = -5} $

---

4.

$$
\frac{-2}{x - 1} = \frac{x - 8}{x + 1}
$$

Step 1: Cross-multiply:
$$
-2(x + 1) = (x - 8)(x - 1)
$$

Step 2: Expand both sides:

Left: $ -2x - 2 $

Right: $ x(x - 1) - 8(x - 1) = x^2 - x - 8x + 8 = x^2 - 9x + 8 $

So:
$$
-2x - 2 = x^2 - 9x + 8
$$

Step 3: Bring all terms to one side:
$$
0 = x^2 - 9x + 8 + 2x + 2 = x^2 - 7x + 10
$$

Step 4: Factor:
$$
x^2 - 7x + 10 = (x - 5)(x - 2) = 0
\Rightarrow x = 5 \quad \text{or} \quad x = 2
$$

Check restrictions:
- $ x - 1 \neq 0 \Rightarrow x \neq 1 $
- $ x + 1 \neq 0 \Rightarrow x \neq -1 $

Both $ x = 5 $ and $ x = 2 $ are valid.

✔ Solutions: $ \boxed{x = 5} $ or $ \boxed{x = 2} $

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✔ Final Answers:

1. $ x = -\dfrac{1}{7} $
2. $ x = 4 $ or $ x = -4 $
3. $ x = 7 $ or $ x = -5 $
4. $ x = 5 $ or $ x = 2 $

Let me know if you'd like a visual explanation or graphing!