[tex]\dfrac{x+1}{x-3} = \dfrac{x-2}{x} \ \ |\cdot x(x-3)\\\\\\x-3 \neq 0 \ \ \wedge \ \ x\neq 0\\x \neq 3 \ \ \wedge \ \ x\neq 0\\D = R \setminus\{0,3\}\\\\\\(x-2)(x-3) = x(x+1)\\\\x^{2}-3x-2x+6 = x^{2}+x\\\\x^{2}-x^{2}-5x-x = -6\\\\-6x = -6 \ \ /:(-6)\\\\\boxed{x = 1}[/tex]
