How do you simplify $\frac{4 x y^{3}}{x^{2} y} \cdot \frac{y}{8 x}$ $(4xy^3)/(x^2y)*y/(8x)$ ?

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Answer 1 · 2018-05-12T11:43:08

Multiply and divide out common factors

$\frac{4 x y^{3}}{x^{2} y} \times \frac{y}{8 x} = \frac{4 x y^{4}}{8 x^{3} y^{1}}$ $(4xy^3) /(x^2y) xx y/(8x) = (4 x y^4)/(8x^3 y^1)$

The common factors are 4xy Divide both the numerator and the denominator by the the common factors of 4xy

$\frac{4 x y^{4}}{4 x y} = y^{3}$ $(4xy^4)/(4xy) = y^3$

$\frac{8 x^{3} y^{1}}{4 x y} = 2 x^{2}$ $(8 x^3y^1)/(4xy)= 2x^2$

So the simplified version of the fraction is

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

How do you simplify 4xy3x2y⋅y8x(4xy^3)/(x^2y)*y/(8x)?