Question

1. solve the equation \\(\frac{18}{x^2 - 3x} - \frac{6}{x - 3} = \frac{5}{x}\\) by using the lcd. select the solution(s).

Explanation:

Step1: Factor the denominator

First, factor \(x^{2}-3x\) as \(x(x - 3)\). So the equation becomes \(\frac{18}{x(x - 3)}-\frac{6}{x - 3}=\frac{5}{x}\).

Step2: Find the LCD

The denominators are \(x(x - 3)\), \(x - 3\), and \(x\). The least common denominator (LCD) is \(x(x - 3)\).

Step3: Multiply each term by LCD

Multiply each term in the equation by \(x(x - 3)\) to eliminate the denominators:
\[

$$\begin{align*} x(x - 3)\times\frac{18}{x(x - 3)}-x(x - 3)\times\frac{6}{x - 3}&=x(x - 3)\times\frac{5}{x}\\ 18-6x&=5(x - 3) \end{align*}$$

\]

Step4: Simplify and solve the equation

Expand the right - hand side: \(18-6x = 5x-15\).
Add \(6x\) to both sides: \(18=5x + 6x-15\), which simplifies to \(18 = 11x-15\).
Add 15 to both sides: \(18 + 15=11x\), so \(33 = 11x\).
Divide both sides by 11: \(x = 3\).

Step5: Check for extraneous solutions

We need to check if \(x = 3\) makes the original denominators zero.
For the denominator \(x^{2}-3x=x(x - 3)\), when \(x = 3\), \(x(x - 3)=3\times(3 - 3)=0\).
For the denominator \(x - 3\), when \(x = 3\), \(x - 3=0\).
For the denominator \(x\), when \(x = 3\), \(x=3
eq0\), but since two of the denominators are zero when \(x = 3\), \(x = 3\) is an extraneous solution. So there is no solution.

Answer:

No solution (since \(x = 3\) is extraneous)