$$f\circ g=\frac { \dfrac { x+3 }{ x-6 } -2 }{ \dfrac { x+3 }{ x-6 } +8 }$$
How would I solve this complex fraction? I know what the answer is, but I am just not sure how they got there. The answer is $$\frac{-x+15}{9x-45}$$
I have tried multiplying both sides by $(x-6)$ but I am getting $x^2-3x-18$? What am I doing wrong here?
$$f\circ g=\frac { \frac { x+3 }{ x-6 } -2 }{ \frac { x+3 }{ x-6 } +8 } =\frac { \frac { x+3-2x+12 }{ x-6 } }{ \frac { x+3+8x-48 }{ x-6 } } =\frac { -x+15 }{ x-6 } \cdot \frac { x-6 }{ 9x-45 } =\frac { 15-x }{ 9x-45 } $$