Question 252984
your formula is:


{{{(-5y^4*(y^5)^2) / (15y^7*(y^2)^3)}}}


{{{(y^5)^2 = y^(5*2) = y^10}}}


{{{-5*y^4*y^10 = -5*y^(4+10) = -5*y^14}}}


{{{(y^2)^3 = y^(2*3) = y^6}}}


{{{15*y^7*y^6 = 15*y^(7+6) = 15*y^(13)}}}


your equation becomes:


{{{(-5*y^14)/(15*y^13)}}}


{{{y^14/y^13 = y^(14-13) = y}}}


your equation becomes:


{{{(-5*y)/(15)}}} which becomes:


{{{-y/3}}}


to confirm, let y be any value.


try 5.


original equation is:


{{{(-5y^4*(y^5)^2) / (15y^7*(y^2)^3)}}}


when y = 5, this equation becomes:


{{{(-5*(5)^4*((5)^5)^2) / (15*(5)^7*((5)^2)^3)}}}


this becomes:


{{{-3.051757813^10 / 1.831054688^10 = -1.6666666667 = -5/3}}}


final equation is:


{{{-y/3}}}


this becomes:


{{{-(5)/3}}}


the answers are the same confirming the final equation is equivalent to the original equation.