Answer:
[tex]x=\dfrac{42}{13}[/tex]
Step-by-step explanation:
Given equation:
[tex]\dfrac{6}{(x^2-3x)}=\dfrac{12}{x}+\dfrac{1}{(x-3)}[/tex]
Factor the denominator on the left side:
[tex]\implies \dfrac{6}{x(x-3)}=\dfrac{12}{x}+\dfrac{1}{(x-3)}[/tex]
Combine the fractions on the right side by multiplying by the LCM:
[tex]\begin{aligned}\implies \dfrac{6}{x(x-3)} & =\dfrac{12}{x} \cdot \dfrac{(x-3)}{(x-3)}+\dfrac{1}{(x-3)} \cdot \dfrac{x}{x}\\\\& =\dfrac{12(x-3)}{x(x-3)}+\dfrac{x}{x(x-3)}\\\\& =\dfrac{12(x-3)+x}{x(x-3)} \end{aligned}[/tex]
As the denominators are now the same on both sides:
[tex]\begin{aligned}\implies \dfrac{6}{x(x-3)} & =\dfrac{12(x-3)+x}{x(x-3)}\\\\ \implies 6 & = 12(x-3)+x\end{aligned}[/tex]
Simplify and solve for x:
[tex]\begin{aligned}\implies 6 & = 12(x-3)+x\\\\6 & =12x-36+x\\\\6 & =13x-36\\\\13x & =42\\\\\implies x & =\dfrac{42}{13}\end{aligned}[/tex]
