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How would I go on about solving this:

${x^4y^7\over x^5y^5}$

When $x = {1\over3}$ and y = ${2\over 9}$

My working out:

Firstly I simplify.-

${xy^2\over x}$

Then substitute

${{{1\over3} * {2\over9}}^2\over{1\over3}}$


${{{1\over3} * {4\over81}}\over{1\over3}}$


${{4\over243} \over{1\over3}}$

since $a/b / c/d = ab * dc$:-

${4\over243} * {3\over1}$





The correct answer is


Can someone help me employ the proper method in solving this problem?


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The initial simplification should be to $\frac{y^2}{x}$. – André Nicolas Feb 5 '14 at 5:14
up vote 1 down vote accepted

The first step is $$\frac{y^2}{x}$$ So the answer is $$ \frac{4/81}{1/3}= \frac{4 \cdot 3}{81} = \frac{4}{27}$$

You just had the extra $x$. Must be an oversight

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Argh..9 hours of re-learning maths causes this to happen. Thanks! Thought I was going insane. – User404 Feb 5 '14 at 5:16

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