Question

How do you divide `\frac { r ^ { 2} - 3r - 18} { 3r ^ { 2} - 18r } \div \frac { r + 9} { r ^ { 2} + 12r + 27}`?

Answer 1

First get the reciprocal of `(r+9)/(r^2+12r+27)` and then multiply it to `(r^2 - 3r - 18)/(3r^2- 18r)`

Like so...

`(r^2 - 3r - 18)/(3r^2- 18r)` * `(r^2+12r+27)/(r+9)`

Factor out each of the components to make solving easier

`((r+3)(r-6))/(3r(r-6))` * `((r+3)(r+9))/((r+9))`

Simplify

`((r+3)cancel(r-6))/((3r)cancel(r-6))` * `((r+3)cancel(r+9))/(cancel(r+9)`

Simplify once more to get the final answer

`(r+3)^2/(3r)`

Answer 2

`=1/3+2/r+3/r^2`

Explanation:

(1) Use the rule of dividing fractions .

That is invert second fraction and change the problem to multiplication

`\frac { r ^ { 2} - 3r - 18} { 3r ^ { 2} - 18r } \div \frac { r + 9} { r ^ { 2} + 12r + 27}`?

`(r^2-3r-18)/(3r^2-18r)xx(r^2+12r+27)/(r+9)`

2) Factorise fully as possible

`((r-6)(r+3))/(3r(r-6))xx((r+3)(r+9))/(r+9)`

(3) Cancel down

`(cancel((r-6))(r+3))/(3rcancel((r-6)))xx((r+3)cancel((r+9)))/cancel((r+9))`

(4) Tidy up

`((r+3)^2)/(3r)`

`=(r^2+6r+9)/(3r^2)`

`=1/3+2/r+3/r^2`