How do you simplify $\frac{x^{2}}{x y^{- 4}} \cdot \frac{2 x^{- 3} y^{4}}{3 x y^{- 1}}$ $( x^2 )/(xy^-4)*(2x^-3y^4)/( 3xy^-1)$ ?

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Answer 1 · 2015-06-03T22:08:10

$\frac{2 y^{9}}{3 x^{3}}$ $(2y^9)/(3x^3)$

$\frac{x^{2}}{x y^{- 4}} \cdot \frac{2 x^{- 3} y^{4}}{3 x y^{- 1}}$ $(x^2)/(xy^-4) * (2x^-3y^4)/(3xy^-1)$

We start by moving the negative exponents into positive reciprocals.

$\frac{x^{2} y^{4}}{x} \cdot \frac{2 y^{5}}{3 x^{4}}$ $(x^2y^4)/x * (2y^5)/(3x^4)$

We can cancel a $x^{2}$ $x^2$ from the numerator and denominator.
$\frac{y^{4}}{x} \cdot \frac{2 y^{5}}{3 x^{2}}$ $y^4/x * (2y^5)/(3x^2)$

$\frac{2 y^{9}}{3 x^{3}}$ $(2y^9)/(3x^3)$

How do you simplify ( x^2 )/(xy^-4)*(2x^-3y^4)/( 3xy^-1)x2xy−4⋅2x−3y43xy−1( x^2 )/(xy^-4)*(2x^-3y^4)/( 3xy^-1)?