How do you simplify #\frac { 16y ^ { 4} z ^ { 4} } { 8y z ^ { 6} }#?

1 Answer
G_Ozdilek
Jun 15, 2017

Follow explanation. #(2timesy^3)/z^2#

Explanation:

You can write down your equation:

#(8times2timesytimesy^3timesz^4)/(8timesytimesz^4timesz^2)#

You can see that #y# (on the numerator and the denumerator) and #z^4# on both can be cancelled out as well as 8, yielding

#(2timesy^3)/z^2#

This is your answer:

#(2timesy^3)/z^2#

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