Question

How do you multiply `(18x-36)/(4x-8)*2/(9x+18)` and state the excluded values?

Answer 1

=`1/(x+2)`

`x !=2 and x != -2`

Explanation:

`(18x-36)/(4x-8)*2/(9x+18)" "larr` factorise first

`=(18(x-2) xx2)/(4(x-2)xx9(x+2))`

`=(cancel18^cancel2cancel{(x-2)} xx cancel2)/(cancel4cancel{(x-2)}xxcancel9(x+2))" "larr` cancel like factors

`= 1/(x+2)`

Remember that the denominator may not be zero,
The excluded values are therefore the values which
would give us 0.