[tex]\displaystyle\bf\\\frac{x-3}{x^2-9} + \frac{x}{x-3} = 1\\\\\frac{x-3}{x^2-3^2} + \frac{x}{x-3} = 1\\\\\frac{x-3}{(x-3)(x+3)} + \frac{x}{x-3} = 1\\\\\frac{x-3}{(x-3)(x+3)} + \frac{x(x+3)}{(x-3)(x+3)} = 1\\\\\frac{x-3}{x^2-3^2} + \frac{x^2+3x}{x^2-3^2} = 1\\\\\frac{x^2+3x+x-3}{x^2-9}=1\\\\\frac{x^2+4x-3}{x^2-9}=1\\\\x^2+4x-3=x^2-9\\\\x^2+4x-x^2=-9+3\\\\4x=-6\\\\x=\frac{-6}{4}\\\\\boxed{\bf x=-\frac{3}{2}}[/tex]
