Question 103489: Can someone offer some assistance to a 49 year old algebra student?
([2a^2b]/[3x^2y])^3 times ([-6xy^2]/[a^3b])^2 Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! ([2a^2b]/[3x^2y])^3 times ([-6xy^2]/[a^3b])^2
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=([8a^6b^3] / [27x^6y^3]) * ([36x^2y^4] / [a^6b^2])
Get the ab's together and the xy's together:
=([8a^6b^3] / [a^6b^2]) * ([36x^2y^4] / [27x^6y^3])
Cancel where you can to get:
= (8b)*(4y/3x^4)
= [32by/3x^4]
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Cheers,
Stan H.
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