How do you simplify $\frac{4 x y^{3}}{- 2 x^{4} y^{2}}$ $\frac{4x y^{3}}{- 2x^{4} y^{2}}$ ?

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Answer 1 · 2018-06-14T18:34:07

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

Let's deal with the constants first. We'll get

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

The key realization here is that since we have variables in the numerator and the denominator, we can subtract the exponents to get the new exponent.

Our expression can be rewritten as

$- 2 x^{1 - 4} y^{3 - 2}$ $-2x^(1-4)y^(3-2)$

As you see, we just subtract the exponents of the like variables. This simplifies to

$- 2 x^{- 3} y$ $-2x^-3y$

We can make the negative exponent positive by bringing it to the denominator. We now have

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

Hope this helps!

How do you simplify \frac{4x y^{3}}{- 2x^{4} y^{2}}4xy3−2x4y2\frac{4x y^{3}}{- 2x^{4} y^{2}}?