To solve the equation
[
\frac{2x + 8}{x - 2} + 6 = 0,
]
we can start by isolating the fraction. First, subtract 6 from both sides:
[
\frac{2x + 8}{x - 2} = -6.
]
Next, we can eliminate the fraction by multiplying both sides by (x - 2), but we need to ensure that (x \neq 2) because that would make the denominator zero. Multiplying both sides gives:
[
2x + 8 = -6(x - 2).
]
Expanding the right side:
[
2x + 8 = -6x + 12.
]
Now, we collect all the (x) terms on one side and the constant terms on the other side. Adding (6x) to both sides results in:
[
2x + 6x + 8 = 12,
]
or
[
8x + 8 = 12.
]
Next, we subtract 8 from both sides:
[
8x = 4.
]
Dividing both sides by 8, we find:
[
x = \frac{1}{2}.
]
Thus, the solution to the equation is
[
\boxed{\frac{1}{2}}.
]
To verify, we can substitute (x = \frac{1}{2}) back into the original equation:
[
\frac{2\left(\frac{1}{2}\right) + 8}{\frac{1}{2} - 2} + 6 = \frac{1 + 8}{\frac{1}{2} - 2} + 6 = \frac{9}{-\frac{3}{2}} + 6 = -6 + 6 = 0.
]
The equation holds true, confirming (x = \frac{1}{2}) is the correct solution.
