[tex]\dfrac{15}{|x-3|+|4x-12|}=8\\\\\\\dfrac{15}{|x-3|+4|x-3|}=8\\\\\\\dfrac{15}{5|x-3|}=8\\\\\\\dfrac{3}{|x-3|}=8\\\\\\\text{D}\colon\\\\x-3\neq0\\\\x\neq3\\\\\text{D}=\mathbb{R}\setminus\{3\}\\\\\\8|x-3|=3\\\\|x-3|=\dfrac{3}{8}\\\\x-3=\dfrac{3}{8}\quad\text{lub}\quad x-3=-\dfrac{3}{8}\\\\x=\dfrac{27}{8}\quad\text{lub}\quad x=\dfrac{21}{8}\\\\\boxed{x\in\Big\{\dfrac{21}{8},\dfrac{27}{8}\Big\}}[/tex]
