3ab + 5/2b. The book I have states the answer is 6ab^2 + 5/2b. How was the "6" derived? Is it simply 3ab * 2b that brings about 6ab^2?
Thanks!
Well what i meant is, that since you are not given the denominator for 3ab you can assume that it is a fraction with denominator 1, so that the lcm(1,2b)=2b.
Then you can simply multiply 3ab by 2b and 1 by 2b to get a common denominator for 3ab and 5 that is 2b.
So the answer will turn out to be (6ab^2 +5)/2b.
$\displaystyle 3ab+\frac{5}{2b}$
$\displaystyle (\frac{3ab}{1}\times\frac{2b}{2b})+\frac{5}{2b}$
$\displaystyle \frac{6ab^2}{2b}+\frac{5}{2b}$
$\displaystyle \frac{6ab^2+5}{2b}$
I believe that this is correct, but I could be wrong. Someone check my math and verify that it is correct. I don't want to give them the incorrect process. I cannot remember if you can get rid of the b on the bottom or not...