To solve the equation:
[tex]\[
\frac{1}{x+6} + \frac{2}{x+3} = \frac{-3}{x^2 + 9x + 18}
\][/tex]
First, let's simplify and rewrite the denominator on the right-hand side of the equation:
[tex]\[
x^2 + 9x + 18 = (x+3)(x+6)
\][/tex]
So, the equation becomes:
[tex]\[
\frac{1}{x+6} + \frac{2}{x+3} = \frac{-3}{(x+3)(x+6)}
\][/tex]
Next, let's find a common denominator for the fractions on the left-hand side. The common denominator is [tex]\((x+3)(x+6)\)[/tex]. Rewrite each fraction with this common denominator:
[tex]\[
\frac{1 \cdot (x+3)}{(x+6)(x+3)} + \frac{2 \cdot (x+6)}{(x+3)(x+6)} = \frac{-3}{(x+3)(x+6)}
\][/tex]
Simplify the numerators:
[tex]\[
\frac{x + 3}{(x+6)(x+3)} + \frac{2x + 12}{(x+6)(x+3)} = \frac{-3}{(x+3)(x+6)}
\][/tex]
Combine the fractions on the left-hand side:
[tex]\[
\frac{(x+3) + (2x+12)}{(x+3)(x+6)} = \frac{-3}{(x+3)(x+6)}
\][/tex]
Simplify the numerator:
[tex]\[
\frac{3x + 15}{(x+3)(x+6)} = \frac{-3}{(x+3)(x+6)}
\][/tex]
Since the denominators are the same, equate the numerators:
[tex]\[
3x + 15 = -3
\][/tex]
Solve for [tex]\(x\)[/tex]:
[tex]\[
3x + 15 = -3
\][/tex]
Subtract 15 from both sides:
[tex]\[
3x = -3 - 15
\][/tex]
[tex]\[
3x = -18
\][/tex]
Divide by 3:
[tex]\[
x = -6
\][/tex]
Now, let's check if [tex]\(x = -6\)[/tex] is a valid solution by substituting back into the original equation. We notice that if [tex]\(x = -6\)[/tex], the terms [tex]\(\frac{1}{x+6}\)[/tex] become undefined:
[tex]\[
\frac{1}{-6+6} = \frac{1}{0}
\][/tex]
which is undefined. Therefore, [tex]\(x = -6\)[/tex] is not a valid solution.
Given the provided choices:
A. [tex]\(\{0\}\)[/tex]
B. [tex]\(\{-6\}\)[/tex]
C. [tex]\(\{3\}\)[/tex]
D. [tex]\(\varnothing\)[/tex]
None of the choices [tex]\(0, -6, 3\)[/tex] serve as valid solutions, which means the correct answer is:
[tex]\[
\text{D. } \varnothing
\][/tex]
So, the solution set is an empty set:
[tex]\[
\boxed{\varnothing}
\][/tex]
