Which of the following would be an extraneous solution to the following equation if it's a solution? $$ \frac{x+8}{x-2}+\frac{6}{x}=\frac{x-4}{x} $$ -8 4 6 2

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We start with the equation

$$
\frac{x+8}{x-2} + \frac{6}{x} = \frac{x-4}{x}
$$

**Step 1. Identify the domain restrictions.**

The denominators $$ x-2 $$ and $$ x $$ cannot be zero. Thus,

$$
x-2 \neq 0 \implies x \neq 2,\quad x \neq 0.
$$

Any solution that produces $$ x = 2 $$ or $$ x = 0 $$ would be extraneous.

**Step 2. Multiply by the common denominator $$ x(x-2) $$ to eliminate the fractions.**

Multiplying every term by $$ x(x-2) $$:

$$
x(x-2) \left( \frac{x+8}{x-2} \right) + x(x-2) \left( \frac{6}{x} \right) = x(x-2) \left( \frac{x-4}{x} \right)
$$

Simplify each term:

- For the first term:
  $$
  x(x-2) \cdot \frac{x+8}{x-2} = x(x+8)
  $$
  
- For the second term:
  $$
  x(x-2) \cdot \frac{6}{x} = 6(x-2)
  $$
  
- For the third term:
  $$
  x(x-2) \cdot \frac{x-4}{x} = (x-4)(x-2)
  $$

The equation becomes

$$
x(x+8) + 6(x-2) = (x-4)(x-2)
$$

**Step 3. Expand and simplify the equation.**

Expand the left-hand side:

$$
x^2 + 8x + 6x - 12 = x^2 + 14x - 12
$$

Expand the right-hand side:

$$
(x-4)(x-2) = x^2 - 2x - 4x + 8 = x^2 - 6x + 8
$$

Now the simplified equation is:

$$
x^2 + 14x - 12 = x^2 - 6x + 8
$$

Subtract $$ x^2 $$ from both sides:

$$
14x - 12 = -6x + 8
$$

Add $$ 6x $$ to both sides:

$$
20x - 12 = 8
$$

Now add $$ 12 $$ to both sides:

$$
20x = 20
$$

Divide by 20:

$$
x = 1
$$

**Step 4. Identify the extraneous solution among the given choices.**

The only candidate solution from the algebraic manipulation is $$ x = 1 $$, and it does not violate the domain restrictions since $$ 1 \neq 0 $$ and $$ 1 \neq 2 $$.

The question, however, asks: "Which of the following would be an extraneous solution to the equation if it's a solution?" This means we must identify any candidate that would be excluded due to the domain restrictions.

Among the options $$-8$$, $$4$$, $$6$$, and $$2$$, notice that:

- $$ x = 2 $$ is not allowed because it would make the denominator $$ x-2 $$ zero.

Thus, $$ x = 2 $$ would be an extraneous solution.

$$
\boxed{2}
$$
$$ x = 2 $$ is an extraneous solution.

Which of the following would be an extraneous solution to the following equation if it's a solution? \( \frac{x+8}{x-2}+\frac{6}{x}=\frac{x-4}{x} \) -8 4 6 2

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