Question 194422: Complex Fraction
Question: ab + b^2 / 4ab^5 divided by a+b/6a^2b^4 =
I was lost on the last example so I am lost again on this one.
Please explain the way to get through this problem. Thank You.
Answer by ptaylor(2198)   (Show Source):
You can put this solution on YOUR website!
OK. Complex fractions can be readily solved by first making the denominator of the complex fraction (in this case, (a+b)/6a^2b^4) equal to 1. We can do this by multiplying it by 6a^2b^4/(a+b). When we do that, however, we have to also multiply the numerator of the complex fraction (in this case,(ab+b^2)/4ab^5) by 6a^2b^4/(a+b). Why? Because 6a^2b^4/(a+b)/6a^2b^4/(a+b) is equal to 1 and we do not change the complex fraction by multiplying it by 1.
Now, lets take a simple example: (a/b)/(c/d). Multiply both the numberator and denominator by (d/c) and we get:(a/b)(d/c)/(c/d)(d/c)= (a/b)(d/c)/1=ad/bc.
Now, in your problem:
a=(ab+b^2)=b(a+b)
b=4ab^5
c=(a+b)
d=6a^2b^4
and
ad/bc=b(a+b)*6a^2b^4/4ab^5*(a+b)=
b*6a^2b^4/4ab^5 =6a^2b^5/4ab^5=(3/2)a ---is that correct???
Hope this helps---ptaylor
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