SOLUTION: I am having trouble remembering how to start complex fractions. I know to multiply by the inverse, but it is ot making any sence to me. Here is the Expression {{{(3+1/3b)/(2-1/6b)}

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: I am having trouble remembering how to start complex fractions. I know to multiply by the inverse, but it is ot making any sence to me. Here is the Expression {{{(3+1/3b)/(2-1/6b)}      Log On


   

Question 135618: I am having trouble remembering how to start complex fractions. I know to multiply by the inverse, but it is ot making any sence to me. Here is the Expression %283%2B1%2F3b%29%2F%282-1%2F6b%29. Thank you for your time in helping me with this troublesome expression.
Found 2 solutions by solver91311, stanbon:
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
The first thing you need to do is simplify both the numerator and the denominator by finding and applying the lowest common denominator in each.



Now you can multiply the numerator by the inverse (By the way that is more properly termed the multiplicative inverse or reciprocal to avoid confusion with the additive inverse)

%28%289b%2B1%29%2F%283b%29%29%286b%2F%2812b-1%29%29

You should be able to take it from here. Remember to take out the common factor of 3b.





Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
(3+(1/3b))/(2-(1/6b))
= [(9b+1)/3b] / [(12b-1)/6b]
Invert the denominator and multiply to get:
= [(9b+1)/3b] / [6b/(12b-1)]
= [2(9b+1)/(12b-1)]
= [24b-2]/[12b-1]
=====================
Cheers,
Stan H.