Answer :

[tex]\frac{96}{126}\\\\96\ and\ 126\ are\ even\ numbers\ divisible\ by\ 2\\\\\frac{96:2}{126:2}=\frac{48}{63}\\\\48\ and\ 63\ are\ divisible\ by\ 3,\ because\ 4+8=12(divisible\ by\ 3)\\and\ 6+3=9(divisible\ by\ 3)\\\\\frac{48:3}{63:3}=\boxed{\frac{16}{21}}\leftarrow answer[/tex]
thompzac
16/21 because like I had already said before 6 can go into each of theses 16 times and 21 times and 16/21 is the lowest possible that you could go with this because 21 is non divisible by a any whole number that 16 is.