Questions: 18/(x^2-3x)-6/(x-3)=5/x

Solution Steps

We start with the equation

\[ \frac{18}{x^{2}-3x}-\frac{6}{x-3}=\frac{5}{x}. \]

First, we factor the polynomial in the denominator of the first term, \(x^{2}-3x\):

\[ x^{2}-3x = x(x - 3). \]

Next, we rewrite the left side of the equation using partial fraction decomposition. We express

\[ \frac{18}{x(x - 3)} = \frac{A}{x} + \frac{B}{x - 3}. \]

After performing the decomposition, we find:

\[ \frac{18}{x(x - 3)} = \frac{6}{x - 3} - \frac{6}{x}. \]

Now, we substitute the partial fraction decomposition back into the original equation:

\[ \frac{6}{x - 3} - \frac{6}{x} - \frac{6}{x - 3} = 0. \]

This simplifies to:

\[ -\frac{6}{x} = 0. \]

The equation

\[ -\frac{6}{x} = 0 \]

implies that there are no solutions for \(x\) since the left-hand side cannot equal zero for any real value of \(x\). Thus, the solution set is empty:

\[ \text{Solution: } \emptyset. \]

Final Answer