Simplifying

**johntuan** · September 11th 2008, 06:51 PM

Sorry for all the questions but i cant seem to figure this one out...

The questions is
[(x^-3)(y^-1)] + y^-5 / [(x^-4)(y5)]

The answer is (xy^4)+(x^4) / y^10

**Shyam** · September 11th 2008, 07:06 PM

Originally Posted by johntuan

Sorry for all the questions but i cant seem to figure this one out...

The questions is
[(x^-3)(y^-1)] + y^-5 / [(x^-4)(y5)]

The answer is (xy^4)+(x^4) / y^10

Is your question like this?

$<br /> x^{-3}y^{-1} + \frac{y^{-5}} {x^{-4}y^5}$

Is your answer like this?

$xy^4 + \frac{x^4}{y^{10}}$ or it is $\frac{xy^4 + x^4}{y^{10}}$

**johntuan** · September 11th 2008, 07:10 PM

Originally Posted by Shyam

Is your question like this?

$<br /> x^{-3}y^{-1} + \frac{y^{-5}} {x^{-4}y^5}$

Is your answer like this?

$xy^4 + \frac{x^4}{y^{10}}$ or it is $\frac{xy^4 + x^4}{y^{10}}$

for the question its
[(x^-3)(y^-1)] + y^-5] / [(x^-4)(y5)], so basically its all over [(x^-4)(y5)].

and for the answer its the second one...sorry for the mixup.

**Shyam** · September 11th 2008, 07:21 PM

Originally Posted by johntuan

Sorry for all the questions but i cant seem to figure this one out...

The questions is
[(x^-3)(y^-1)] + y^-5 / [(x^-4)(y5)]

The answer is (xy^4)+(x^4) / y^10

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

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

$=\left(\frac{1}{x^3y}+\frac{1}{y^5}\right)\div ({x^{-4}y^5})$

$= \left(\frac {y^5+x^3y}{x^3y^6}\right)\times \frac{1}{x^{-4}y^5}$

$= \frac {y(y^4+x^3)}{x^3y^6}\times \frac{x^4}{y^5}$

$= \frac {y^4+x^3}{y^5}\times \frac{x}{y^5}$

$= \frac {xy^4+x^4}{y^{10}}$

Did you get it NOW ???

**johntuan** · September 11th 2008, 07:25 PM

Originally Posted by Shyam

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

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

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

$= \frac {y^4+x^3}{x^3y^5}\times \frac{x^4}{y^5}$

$= \frac {y^4+x^3}{y^5}\times \frac{x}{y^5}$

$= \frac {xy^4+x^4}{y^{10}}$

i'm still a bit confused how did u get from the 3rd step to the 4th?

**Shyam** · September 11th 2008, 07:59 PM

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

$=\left(\frac{1}{x^3y}+\frac{1}{y^5}\right)\div ({x^{-4}y^5})$

(inside the brackets, for numerators, multiply the Numerator of first with Denominator of second, multiply the Numerator of second with Denominator of first), (and for denominator, multiply both the denominators)

$= \left(\frac {1\times y^5+x^3y\times 1}{x^3y \times y^5}\right)\times \frac{1}{x^{-4}y^5}$

$= \left(\frac {y^5+x^3y}{x^3y^6}\right)\times \frac{1}{x^{-4}y^5}$

(take y common from numerator)

$= \frac {y(y^4+x^3)}{x^3y^6}\times \frac{x^4}{y^5}$

$= \frac {y^4+x^3}{y^5}\times \frac{x}{y^5}$

$= \frac {xy^4+x^4}{y^{10}}$

Do you know how to add fractions ??? did you get it now??

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