ATAR Notes: Forum

VCE Stuff => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematics => Topic started by: crayolé on April 07, 2010, 05:24:10 pm

Title: How can I get more practice with questions like these?
Post by: crayolé on April 07, 2010, 05:24:10 pm: Sometimes I have trouble with the fundamental rules of fractions/indicies etc when algebra is involved

I dont know how to explain it but for example,

$(\frac{x^-2 + y^4 + z^5}{x^5 + y^3 + z^-3})^-1$ <---thats everything in the bracket to the power of -1

can be flipped upside down or something?

Is there a list of different rules somewhere?
Title: Re: How can I get more practice with questions like these?
Post by: the.watchman on April 07, 2010, 05:42:32 pm: In general, $k^{-1} = \frac{1}{k}$ (by definition)

So the simplification of your expression is its reciprocal, $\frac{x^5+y^2+z^{-3}}{x^{-2}+y^4+z^5}$

It would be handy for you to note one of the index laws: $k^{-p} = \frac{1}{k^p}$
Title: Re: How can I get more practice with questions like these?
Post by: ReVeL on April 07, 2010, 05:47:16 pm: Remember the index laws.

$x^{-1}= \frac{1}{x}$

Therefore,

$(\frac{x^-2 + y^4 + z^5}{x^5 + y^3 + z^-3})^{-1}=\frac{1}{\frac{x^-2 + y^4 + z^5}{x^5 + y^3 + z^-3}}$

And 1 divided by a fraction is the same as flipping that fraction. Hence,

$\frac{1}{\frac{x^-2 + y^4 + z^5}{x^5 + y^3 + z^-3}}=\frac{x^5 + y^3 + z^-3}{x^-2 + y^4 + z^5}$

But yeah, get familar with the index laws.

EDIT: Beaten. Above explanation is probably better.
Title: Re: How can I get more practice with questions like these?
Post by: Blakhitman on April 07, 2010, 05:48:45 pm: http://www.staff.vu.edu.au/mcaonline/units/indices/indexlaws.html

Got practice questions too lol.
Title: Re: How can I get more practice with questions like these?
Post by: GerrySly on April 07, 2010, 06:07:14 pm: $\begin{align*} \left (\frac{x^{-2}+y^4+z^5}{x^5+y^3+z^{-3}}\right )^{-1}&=\left (\frac{\frac{1}{x^2}+y^4+z^5}{x^5+y^3+\frac{1}{z^3}}\right )^{-1}\text{ (Since }a^{-b}=\frac{1}{a^b})\\ &=\left (\frac{\frac{1+x^2y^4+x^2z^5}{x^2}}{\frac{x^5z^3+y^3z^3+1}{z^3}}\right )^{-1}\text{ (Since }\frac{1}{a^b}+x=\frac{1+x\cdot a^b}{a^b})\\ &=\left (\frac{(1+x^2y^4+x^2z^5)(z^3)}{(x^5z^3+y^3z^3+1)(x^2)}\right )^{-1}\text{ (Since }\frac{\frac{a}{b}}{\frac{c}{d}}=\frac{a\cdot d}{b\cdot c})\\ &=\left (\frac{z^3+x^2y^4z^3+x^2z^8}{x^7z^3+y^3z^3x^2+x^2}\right )^{-1}\text{ (Since }x^2\cdot x^7=x^9)\\ &=\frac{x^7z^3+y^3z^3x^2+x^2}{z^3+x^2y^4z^3+x^2z^8}\text{ (Since }\left (\frac{a}{b}\right )^{-1}=\frac{1}{\frac{a}{b}}=\frac{b}{a}) \end{align*}$

Well I was bored and thought I'd go through the entire thing, you wouldn't do line 3 but as I said, I'm pretty bored