Question 384603
I need help dividing the following: 
[3y+12/8y^3] ÷ [9y+36/16y^3]
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[3(y+4)/8y^3] *[16y^3/9(y+4)]
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Cancel all factors common to a numerator and a denominator to get:
[1/1] *[2/3]
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= 2/3
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cheers,
Stan H.