Question

(1/b - 1/9) / b-9

Answer 1

[(9-b)/9b]/b-9

Answer 2

(9-b)/9b(b-9)
taking - common,
-(b-9)/9b(b-9)
=-1/9b

Answer 3

\[\frac{ \frac{ 1 }{ b } - \frac{ 1 }{ 9 }}{ b-9}\]

Take LCM of b and 9 in denominator [Which is 9b]

\[\frac{ \frac{ 9-b }{ 9b } }{ b-9 } = \frac{ 9-b }{ 9b(b-9) }\]

NOW TAKE -1 common in numerator.

\[\frac{ -1(b-9) }{ 9b(b-9)} Now~~Cancel~~(b-9)\]

You'd be left with.

\[\frac{ -1 }{ 9b }\]

Understood this? :)

Answer 4

I understood using the Least common multiplier, but missed how you made 1/b-1/9 = 9-b/9b?

Answer 5

To add ANY fraction we need a common denominator.

1/b - 1/9 = 9/9b - b/9b
[Multiply first fraction with 9 to get denominator 9b and multiply the second fraction with b to get denominator 9b]

So
9/9b - b/9b
(9-b)/9b

Now Understood? :)

Answer 6

Oh Man, Yes, forgot the basic rules, Thank you, I think I will stop after this question and gets some sleep, is almost 1am here, Thank you VERY much

Answer 7

:)