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Dawny521 · Jun 16, 2009, 04:58 PM

Simplify express ans./only positive exponents

(x^-2y^4)/(4y^-3)^2

=1/4x^-2^-2y^4(-3)
=1/4x^-2-2y^4+3
=1/4x^4y^1

=y/4x^4
OK now this is wrong from what I am told but I don't know why and how its wrong can some one show me how this is done please

Perito · Jun 16, 2009, 08:08 PM

$(x^{-2}y^4)/(4y^{-3})^2$

I'm not positive I have the final exponent (2) in the right place, but it's where you wrote it, so I'll go from here.

$x^{-2} = \frac {1}{x^2}$

$\frac {1}{y^{-3}}=y^3$

$=\Large \frac {\left ( \frac {y^4}{x^2} \right)}{ \left( \frac {4}{y^3} \right)^2}$

$=\Large \frac {\left ( \frac {y^4}{x^2} \right)}{ \left( \frac {16}{y^6} \right)}= \frac {y^4}{x^2}\times \frac {y^6}{16}=\frac {y^{10}}{16x^2}$