# How to solve complex fractions?

$$f\circ g=\frac { \dfrac { x+3 }{ x-6 } -2 }{ \dfrac { x+3 }{ x-6 } +8 }$$

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?

• you mean $f\circ g=\frac { x+3 }{ x-6 } -\frac { 2 }{ \frac { x+3 }{ x-6 } } +8$? – haqnatural Jul 29 '16 at 14:57
• Could you please write this out in the latex/mathjax otherwise it is rather difficult to see what you mean. eg is $1+x/3+9/x = \frac{1+x}{3+\frac{9}{x}}$ or $1+\frac{x}{3}+\frac{9}{x}$? – Kitter Catter Jul 29 '16 at 14:57
• If you mean $f\circ g=\frac { x+3 }{ x-6 } -\frac { 2 }{ \frac { x+3 }{ x-6 } } +8$ , assume$\frac { x+3 }{ x-6 } =t$ and you will get quadratic – Aakash Kumar Jul 29 '16 at 15:01
• I added a photo guys, sorry Im terrible at formatting on this forum. – Trenton Tyler Jul 29 '16 at 15:03
• @TrentonTyler Also add image of question – Aakash Kumar Jul 29 '16 at 15:06

$$f\circ g=\frac { \frac { x+3 }{ x-6 } -2 }{ \frac { x+3 }{ x-6 } +8 } =\frac { \frac { x+3-2x+12 }{ x-6 } }{ \frac { x+3+8x-48 }{ x-6 } } =\frac { -x+15 }{ x-6 } \cdot \frac { x-6 }{ 9x-45 } =\frac { 15-x }{ 9x-45 }$$
• $-6*8 = -48$... – Neil W Jul 29 '16 at 15:12