Question

What is `(5x-10)/(7x+14) * (6x+12)/(14x-28)`?

Answer 1

`15/49`

Explanation:

This is all one term, but we need to have factors before we can cancel. Factorise each part first.

=`(5x-10)/(7x+14) * (6x+12)/(14x-28)`

`(5(x-2))/(7(x+2)) xx (6(x+2))/(14(x-2))`

=`(5cancel(x-2))/(7cancel(x+2)) xx (cancel6^3cancel(x+2))/(cancel14^7cancel(x-2))`

=`15/49`

Answer 2

`15/49`

Explanation:

This expression incorporates the difference of 2 squares but they are in disguise: expression type `a^2-b^2=(a-b)(a+b)`

Write as `(5(x-2))/(7(x+2))xx(6(x+2))/(14(x-2))`

From this point we have two paths. Both versions of the same thing.

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Using the distributive property laws, that is of type : `2xx3=3xx2` we can write:

`(5xx6)/(7xx14)xx(x-2)/(x-2)xx(x+2)/(x+2)`

`30/98xx1xx1 = 15/49`

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Using the difference of 2 squares

`(5xx6xx(x-2)(x+2))/(7xx14xx(x+2)(x-2)) -> (30cancel((x^2-2^2)))/(98(cancel(x^2-2^2))) = 15/49`

`color(red)("I prefer the first method")`