Question 622641
Given to divide:
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{{{ ((3y+12)/(8y^3))/((9y+36)/(16y^3)) }}}
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Note that this is one great big fraction. The numerator is:
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{{{ ((3y+12)/(8y^3))}}}
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and the denominator is:
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{{{((9y+36)/(16y^3)) }}}
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So you are dividing the numerator by the denominator. Remember the rule for dividing by a fraction ... namely that you can divide by a fraction by inverting the fractional divisor and then multiplying that inverted fraction by the number you are dividing into.
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When you invert the denominator you get:
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{{{((16y^3)/(9y + 36)) }}}
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Now multiply that inverted fraction times the numerator and you have:
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{{{((3y+12)/(8y^3))*((16y^3)/(9y + 36)) }}}
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Note that the denominator of the first fraction is {{{8y^3}}} and it will divide into the numerator of the second fraction {{{16y^3}}} to give just 2.  This reduces the problem to:
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{{{((3y+12)/1)*((2)/(9y + 36)) }}}
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Next note that the denominator of the second fraction can be factored by 9 to give:
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{{{((3y+12)/1)*(2/(9(y + 4))) }}}
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Then the numerator of the first fraction can be factored by 3 to result in:
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{{{((3(y+4))/(1))*((2)/(9(y + 4))) }}}
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So you have a (y+4) factor in the numerator and a (y+4) factor in the denominator. These cancel out and you are left with:
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{{{(3/1)*(2/9)}}}
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which after multiplication gives:
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{{{(3*2)/9 = 6/9}}}
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and 6 divided by 9 is equal to {{{2/3}}} 
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and that is the answer to this problem.
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I hope this helps you to see your way through this problem 
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