Question

How do you simplify `((2x-5)/(x^2-9))/((3x-1)/(x+3))`?

Answer 1

`(2x-5)/((x+3)(3x-1))`

Explanation:

Rewrite the problem by moving the bottom fraction in line with the top fraction and adding a division symbol. Rewrite `x^2-9` into `(x-3)(x+3)`:

`(2x-5)/((x-3)(x+3)) -: (3x-1)/(x+3)`

Flip the fraction on the left and rewrite the problem as a multiplication problem:

`(2x-5)/((x-3)(x+3)) xx ((x+3)/(3x-1))`

Cancel out the x+3

`(2x-5)/(x+3) xx (1/(3x-1))`

Consolidate the denominator:

`(2x-5)/((x+3)(3x-1))`

Answer 2

`(2x-5)/(3x^2-10x+3)`

Explanation:

Using an example:

Consider `" "6-:2 = 3`
Although not normally done it is both correct and permissible to write this as: `6/1-:2/1`

The shortcut method is: Turn `2/1` upside down and multiply

`6/1xx1/2 = (6xx1)/(1xx2) = 6/2=3`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Given`" "(2x-5)/(x^2-9)-:(3x-1)/(x+3)`

Write the given expression as: `" "(2x-5)/(x^2-9)xx(x+3)/(3x-1)`

But `x^2-9` is the same as `x^2-3^2=(x-3)(x+3)` giving:

`(2x-5)/((x+3)(x-3))xx(x+3)/(3x-1)`

`(2x-5)/(cancel((x+3))^1(x-3))xx(cancel((x+3))^1)/(3x-1)`

`(2x-5)/((x-3)(3x-1))`

`(2x-5)/(3x^2-10x+3)`