We start with the given equation:
\[ \frac{9}{x-3} = \frac{x^2}{x-3} \]
Since both sides of the equation have the common denominator \(x - 3\), we can rewrite the equation as:
\[ 9 = x^2 \]
To rewrite this as two separate equations joined by "and," we note that the common denominator informs us of a restriction on the variable \(x\):
- \(9 = x^2\) (the equation derived from the original equation)
- \(x - 3 \neq 0\) (we must ensure the denominator is not zero)
The second equation (\(x - 3 \neq 0\)) is significant because it tells us that \(x\) cannot equal 3, to avoid division by zero in the original equation.
Next, we solve the first equation:
\[ 9 = x^2 \]
Taking the square root of both sides gives us:
\[ x = 3 \quad \text{or} \quad x = -3 \]
Now we must consider the restriction from the second equation. The restriction \(x - 3 \neq 0\) tells us that \(x\) cannot be 3. Therefore, we discard the solution \(x = 3\).
The only solution we have that satisfies the equation without violating the restriction is:
\[ x = -3 \]
Interpretation of the Solution Set
The solution set is:
\[ { -3 } \]
This means that the only value of \(x\) that satisfies the original equation while also respecting the restriction (that \(x\) cannot be 3) is \(x = -3\). Thus, -3 is the only valid solution to the equation.