Question 190590
{{{23/(3xy)-1/(3xy^5)}}} Start with the given expression.



{{{(23*y^4)/(3xy*y^4)-1/(3xy^5)}}} Multiply the first fraction by {{{y^4/y^4}}} to get the denominators equal



{{{(23y^4)/(3xy^5)-1/(3xy^5)}}} Multiply



{{{(23y^4-1)/(3xy^5)}}} Combine the fractions.



So {{{23/(3xy)-1/(3xy^5)=(23y^4-1)/(3xy^5)}}} where {{{x<>0}}} or {{{y<>0}}}