[tex]\frac{x+8}{x-2} =\frac{1+4x}{x-2} \\ \\x-2\neq 0 \\x\neq 2 \\D=R\setminus \left \{ 2 \right \}\\\\\frac{x+8}{x-2} -\frac{1+4x}{x-2}=0\\\\\frac{x+8-(1+4x)}{x-2}=0\\\\\frac{x+8- 1-4x }{x-2}=0\\\\\frac{-3x +7}{x-2}=0\\\\-3x+7=0[/tex]
[tex]denominator \ must \ be \ different \ from \ zero, \ so \ the \ counter \ equate \ to \ zero \\\\ -3x+7 =0 \ \ |\ subtract\ 7\ to\ both\ sides \\\\-3x=-7 \ \ | \ divide \ both \ sides\ by\ (-3)\\\\x=\frac{7}{3}[/tex]