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I am doing some pre study/revision in maths before i take my next education steps and am working through a hefty engineering maths book from the beginning. The first chapter is a review of algebraic techniques using indices.

The question is to simplify this

\begin{align} \frac{2xy^2}{(2xy)^2} \end{align}

and the solution given is this

\begin{align} \frac{1}{2x} \end{align}

what are the intermediate steps to arrive at this solution?

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$$\frac{2xy^{2}}{\left( 2xy\right) ^{2}}=\frac{2xy^{2}}{2^{2}x^{2}y^{2}}=% \frac{2xy^{2}/\left( 2xy^{2}\right) }{2^{2}x^{2}y^{2}/\left( 2xy^{2}\right) }% =\frac{1}{2x}$$

provided that $2xy^{2}\neq 0$.

Justification

$\left( 2xy\right) ^{2}=2^{2}x^{2}y^{2}$

$\dfrac{2^{2}x^{2}y^{2}}{2xy^{2}}=2x$

$\dfrac{2xy^{2}}{2xy^{2}}=1$

Alternatively

$$\frac{2xy^{2}}{\left( 2xy\right) ^{2}}=2xy^{2}\left( 2xy\right) ^{-2}=2xy^{2}2^{-2}x^{-2}y^{-2}=2^{1-2}x^{1-2}y^{2-2}=2^{-1}x^{-1}y^{0}=\frac{1}{2x}$$

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  • $\begingroup$ excellent work, thank you,+1 for the double solution, I like the alternative solution using the negative indices rule. $\endgroup$ – Aran Mulholland Nov 10 '10 at 9:54
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Expand (square) the denominator: $$ (2xy)^{2} = 2^{2}\cdot x^{2} \cdot y^{2} = 4x^{2}y^{2}$$

Rewrite the expanded denominator ($4x^{2}y^{2}$) in terms of the numerator ($2xy^{2}$) as its product:

$$ 4x^{2}y^{2} = 2xy^{2} \cdot 2x$$

Thus:

$$ \frac{2xy^{2}}{(2xy)^{2}} = \frac{2xy^{2}}{2x \cdot 2xy^{2}} = \frac{1}{2x} \cdot \frac{2xy^{2}}{2xy^{2}} = \frac{1}{2x} \cdot 1 = \frac{1}{2x} $$

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