Question 74361
{{{4x^2y/2xy^3}}}
{{{4x^2y^1/2x^1y^3}}}When dividing like bases with exponents we subtract the exponents. If I have {{{x^2/x}}} it translates to {{{x*cross(x)/cross(x)=x}}} or {{{x^2/x=x^2/x^1=x^(2-1)=x^1=x}}}. So we're going to apply the same idea for this problem.
{{{4x^(2-1)y^(1-3)/2}}}
{{{4x^1y^(-2)/2}}}Simplify 4/2 to 2
{{{2x^1y^(-2)}}}When we have negative exponents, we invert the base (ie {{{x^(-2)=1/x^2}}} because if we have {{{x/x^3=x^(1-3)=x^(-2)=1/x^2=cross(x)/(cross(x)*x*x)=1/(x*x)=1/x^2}}}. So {{{x^(-a)=1/x^a}}}
{{{(2x^1)/(y^(2))}}}
{{{2x/y^2}}}Here's the simplified form.