Answer :
Let's solve the equation step by step to find the correct number of viable and extraneous solutions along with the viable solution.
Given:
[tex]\[ \frac{3}{x} - \frac{x}{x+6} = \frac{18}{x^2 + 6x} \][/tex]
First, let's rewrite the right-hand side of the equation:
[tex]\[ x^2 + 6x = x(x + 6) \][/tex]
So,
[tex]\[ \frac{18}{x^2 + 6x} = \frac{18}{x(x+6)} \][/tex]
Our equation now becomes:
[tex]\[ \frac{3}{x} - \frac{x}{x+6} = \frac{18}{x(x + 6)} \][/tex]
To clear the denominators, let's find a common denominator which is [tex]\(x(x + 6)\)[/tex]:
[tex]\[ \frac{3(x+6) - x^2}{x(x+6)} = \frac{18}{x(x+6)} \][/tex]
Simplify the numerator on the left-hand side:
[tex]\[ \frac{3x + 18 - x^2}{x(x+6)} = \frac{18}{x(x+6)} \][/tex]
Set the numerators equal to each other:
[tex]\[ 3x + 18 - x^2 = 18 \][/tex]
Simplify further:
[tex]\[ 3x + 18 - x^2 = 18 \][/tex]
[tex]\[ 3x - x^2 = 0 \][/tex]
[tex]\[ x(3 - x) = 0 \][/tex]
So, we have:
[tex]\[ x = 0 \quad \text{or} \quad x = 3 \][/tex]
We must check for extraneous solutions. Substitute back into the original equation:
1. For [tex]\(x = 0\)[/tex]:
[tex]\[ \frac{3}{0} - \frac{0}{0+6} = \frac{18}{0(0+6)} \][/tex]
This results in division by zero, so [tex]\(x = 0\)[/tex] is not a viable solution.
2. For [tex]\(x = 3\)[/tex]:
[tex]\[ \frac{3}{3} - \frac{3}{3+6} = \frac{18}{3(3+6)} \][/tex]
[tex]\[ 1 - \frac{3}{9} = \frac{18}{27} \][/tex]
[tex]\[ 1 - \frac{1}{3} = \frac{2}{3} \][/tex]
[tex]\[ \frac{2}{3} = \frac{2}{3} \][/tex]
This is a valid solution.
Therefore, we have:
- Number of viable solutions: 1
- Number of extraneous solutions: 1 (which is [tex]\(x = 0\)[/tex])
- Viable Solution: [tex]\(x = 3\)[/tex]
Thus, the correct selection in the table is:
\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 & 1 & [tex]$x=3$[/tex] \\
\hline
\end{tabular}
Given:
[tex]\[ \frac{3}{x} - \frac{x}{x+6} = \frac{18}{x^2 + 6x} \][/tex]
First, let's rewrite the right-hand side of the equation:
[tex]\[ x^2 + 6x = x(x + 6) \][/tex]
So,
[tex]\[ \frac{18}{x^2 + 6x} = \frac{18}{x(x+6)} \][/tex]
Our equation now becomes:
[tex]\[ \frac{3}{x} - \frac{x}{x+6} = \frac{18}{x(x + 6)} \][/tex]
To clear the denominators, let's find a common denominator which is [tex]\(x(x + 6)\)[/tex]:
[tex]\[ \frac{3(x+6) - x^2}{x(x+6)} = \frac{18}{x(x+6)} \][/tex]
Simplify the numerator on the left-hand side:
[tex]\[ \frac{3x + 18 - x^2}{x(x+6)} = \frac{18}{x(x+6)} \][/tex]
Set the numerators equal to each other:
[tex]\[ 3x + 18 - x^2 = 18 \][/tex]
Simplify further:
[tex]\[ 3x + 18 - x^2 = 18 \][/tex]
[tex]\[ 3x - x^2 = 0 \][/tex]
[tex]\[ x(3 - x) = 0 \][/tex]
So, we have:
[tex]\[ x = 0 \quad \text{or} \quad x = 3 \][/tex]
We must check for extraneous solutions. Substitute back into the original equation:
1. For [tex]\(x = 0\)[/tex]:
[tex]\[ \frac{3}{0} - \frac{0}{0+6} = \frac{18}{0(0+6)} \][/tex]
This results in division by zero, so [tex]\(x = 0\)[/tex] is not a viable solution.
2. For [tex]\(x = 3\)[/tex]:
[tex]\[ \frac{3}{3} - \frac{3}{3+6} = \frac{18}{3(3+6)} \][/tex]
[tex]\[ 1 - \frac{3}{9} = \frac{18}{27} \][/tex]
[tex]\[ 1 - \frac{1}{3} = \frac{2}{3} \][/tex]
[tex]\[ \frac{2}{3} = \frac{2}{3} \][/tex]
This is a valid solution.
Therefore, we have:
- Number of viable solutions: 1
- Number of extraneous solutions: 1 (which is [tex]\(x = 0\)[/tex])
- Viable Solution: [tex]\(x = 3\)[/tex]
Thus, the correct selection in the table is:
\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 & 1 & [tex]$x=3$[/tex] \\
\hline
\end{tabular}
