Question #a2849

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Question #a2849

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Answer 1 · 2016-07-13T12:27:54

Use some exponent properties to end up with $\frac{x}{y^{2}}$ $x/y^2$ .

Explanation:

The rules of exponents tell us this:
$\frac{x^{a}}{x^{b}} = x^{a - b}$ $x^a/x^b=x^(a-b)$

In other words, when we divide numbers with exponents that have the same base, we subtract the exponents.

In the context of your problem, we will have to apply this property twice. Rewrite the expression as:
$\frac{x^{2} y^{3}}{x y^{5}}$ $(x^2y^3)/(xy^5)$
$\to (\frac{x^{2}}{x}) (\frac{y^{3}}{y^{5}})$ $->(x^2/x)(y^3/y^5)$

Note that $\frac{x^{2}}{x}$ $x^2/x$ is the same thing as $\frac{x^{2}}{x^{1}}$ $x^2/x^1$ ; using the property described above, this simplifies to $x^{2 - 1} = x^{1} = x$ $x^(2-1)=x^1=x$ .

The same process goes for $(\frac{y^{3}}{y^{5}})$ $(y^3/y^5)$ . All we do is subtract the exponents: $y^{3 - 5} = y^{- 2} = \frac{1}{y^{2}}$ $y^(3-5)=y^(-2)=1/y^2$ . Now we have:
$(x) (\frac{1}{y^{2}})$ $(x)(1/y^2)$

This is the same thing as $\frac{x}{y^{2}}$ $x/y^2$ and so we leave that as our final answer. We can't use the subtract the exponents rule on $\frac{x}{y^{2}}$ $x/y^2$ , because they have different bases ( $x$ $x$ and $y$ $y$ ).

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Question #a2849

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