Question 917900
Since 10 and 4 are both multiples of 2,
and 27 and 9 are both multiples of 9,
I would simplify as soon as possible, like this:
{{{(10/9)*(27/4)=10*27/(9*4)=cross(10)^5*cross(27)^3/(cross(9)*cross(4)^2)=5*3/2=15/2}}}
 
Some teachers discourage crossing out, because some students get too enthusiastic and start crossing out willy-nilly, following no logic, but obeying magical rules that they invent. The result is a mistake.
what we really abbreviate by the crossing out is separating factors equal to 1:
{{{10*27/(9*4)=(2*5)*(9*3)/(9*(2*2))=2*5*9*3/(9*2*2)=9*2*5*3/(9*2*2)=(9*2/(9*2))*(5*3/2)=1*(5*3/2)=5*3/2=15/2}}}