Question

[1-(3/x)]/(9-x^2)

Answer 1

\[\large\text{Is this the problem?}\frac{1 - \frac{3}{x}}{9 - x^{2}}\]

Answer 2

\[\large{\frac{1-\frac{3}{x}}{9-x^2}}\]multiply top and bottom by x\[\large{\frac{x-3}{9-x^2}}\]now factorise the denominator\[\frac{x-3}{(3-x)(3+x)}\]and factor out -1 from nominator to make it (3-x)\[\frac{-\cancel{(3-x)}}{\cancel{(3-x)}(3+x)}\]\[=\frac{-1}{3+x}\]

Answer 3

yes thaat was the problem calcmathlete. thank you for explaining lalaly :)