how is the following solved? b/b^2-25 +5/b+5-6/b

Question

how is the following solved?  b/b^2-25 +5/b+5-6/b

owlfred · Answer

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anonymous · Answer

It's not an equation...

anonymous · Answer

do you want to simplify it?

anonymous · Answer

oh you are trying to combine them into one fraction yes?

anonymous · Answer

then the common denominator is b(b+5)(b-5). get everything in that form and combine

anonymous · Answer

$$\frac{b}{b^2-25}+\frac{5}{b+5}-\frac{6}{b}$$

anonymous · Answer

multiply first one top and bottom by b
multiply second one by 
$$(b-5)b$$ and third by $$(b+5)(b-5)$$ then put them over the same denominator and multiply out in the numeator

anonymous · Answer

$$\frac{b\times b + 5\times b(b-5)-6(b+5)(b-5)}{(b-5)(b+5)b}$$

anonymous · Answer

leave the denominator alone, multiply out in the numerator 


$$b^2+5b^2-25b-6b^2-150$$
$$-25b-150$$ is your numerator if i did the algebra correctly

anonymous · Answer

nope i made a mistake!

anonymous · Answer

last term has -6*-25 = 150 not -150 sorry. answer is 
$$\frac{-25b+150}{b(b+5)(b-5)}$$