To find the correct solution for the given equation [tex]\( \frac{3}{x} - \frac{x}{x+6} = \frac{18}{x^2 + 6 x} \)[/tex], let's go through a thorough, step-by-step solution process.
### Step-by-Step Solution:
1. Rewrite the equation:
We start with the given equation:
[tex]\[
\frac{3}{x} - \frac{x}{x+6} = \frac{18}{x^2 + 6x}
\][/tex]
2. Simplify the denominator on the right-hand side:
Notice that [tex]\( x^2 + 6x \)[/tex] can be factored:
[tex]\[
x^2 + 6x = x(x + 6)
\][/tex]
So, the equation becomes:
[tex]\[
\frac{3}{x} - \frac{x}{x+6} = \frac{18}{x(x + 6)}
\][/tex]
3. Combine fractions on the left-hand side:
To combine the fractions on the left-hand side, find a common denominator, which is [tex]\( x(x + 6) \)[/tex]:
[tex]\[
\frac{3(x + 6)}{x(x + 6)} - \frac{x^2}{x(x + 6)}
\][/tex]
Simplify the numerators:
[tex]\[
\frac{3x + 18 - x^2}{x(x + 6)}
\][/tex]
This can be rewritten as:
[tex]\[
\frac{18 + 3x - x^2}{x(x + 6)}
\][/tex]
4. Set the combined form of the left-hand side equal to the right-hand side:
[tex]\[
\frac{18 + 3x - x^2}{x(x + 6)} = \frac{18}{x(x + 6)}
\][/tex]
5. Equate the numerators:
Given that the denominators are the same, we equate the numerators:
[tex]\[
18 + 3x - x^2 = 18
\][/tex]
6. Simplify the resulting equation:
Subtract 18 from both sides:
[tex]\[
3x - x^2 = 0
\][/tex]
Factor the equation:
[tex]\[
x(3 - x) = 0
\][/tex]
This gives us two potential solutions:
[tex]\[
x = 0 \quad \text{or} \quad x = 3
\][/tex]
### Verification of Solutions:
We need to verify which solutions are viable by substituting them back into the original equation:
- Check [tex]\( x = 0 \)[/tex]:
[tex]\[
\frac{3}{0} - \frac{0}{0+6} \quad \text{is undefined (division by zero)}
\][/tex]
Therefore, [tex]\( x = 0 \)[/tex] is an extraneous solution.
- Check [tex]\( x = 3 \)[/tex]:
Substitute [tex]\( x = 3 \)[/tex]:
[tex]\[
\frac{3}{3} - \frac{3}{3+6} = \frac{18}{3^2 + 6 \times 3}
\][/tex]
Simplify:
[tex]\[
1 - \frac{3}{9} = \frac{18}{27}
\][/tex]
Simplify both sides:
[tex]\[
1 - \frac{1}{3} = \frac{2}{3}
\][/tex]
Which is true since:
[tex]\[
\frac{2}{3} = \frac{2}{3}
\][/tex]
Therefore, [tex]\( x = 3 \)[/tex] is a viable solution.
### Conclusion:
From analyzing the steps, we have:
- Number of Viable Solutions: 1
- Number of Extraneous Solutions: 0
- Viable Solution(s): [tex]\( x = 3 \)[/tex]
The correct choice in the table is:
[tex]\[
\begin{tabular}{|c|c|c|}
\hline \begin{tabular}{c}
Number of \\
Viable Solutions
\end{tabular} & \begin{tabular}{c}
Number of \\
Extraneous \\
Solutions
\end{tabular} & Viable Solution(s) \\
\hline
1 & 0 & x = 3 \\
\hline
\end{tabular}
\][/tex]
