Algebra fractions?

I'll use lower case variables so it's easier for me, but the steps are still the same:

{a [1 + (1 / b)] - (2 / b)} / [1 + (1 / b)]

We want to simplify this. So for now, let's ignore the big division and just look at the numerators and denominators separately. Then we can put them back together as a big division problem:

Numerator:

a [1 + (1 / b)] - (2 / b)

Let's start with that sum of fractions in the braces and get the common denominator:

a [b / b + (1 / b)] - (2 / b)

Now we can add the numerators:

a [(b + 1) / b] - (2 / b)

Now let's integrate that "a" as part of the numerator:

a(b + 1) / b - (2 / b)

And now we have difference of two fractions, with the same denominator, so subtract the numerators:

[a(b + 1) - 2] / b

Let's leave that there for now. Next, the denominator of the original problem:

1 + (1 / b)

same as before, common denominator, then add the numerators:

b / b + (1 / b)

(b + 1) / b

Now let's put them back into the big division problem to have the division of two fractions:

{[a(b + 1) - 2] / b} / [(b + 1) / b]

Now we can turn them into the multiplication of the reciprocal:

{[a(b + 1) - 2] / b} * b / (b + 1)

The b's (first denominator and second numerator) cancel out:

[a(b + 1) - 2] / (b + 1)

Now we have the the difference of two expressions in the numerator and a binomial in denominator.

The backwards of adding two fractions (adding the numerators when the denominators are the same), let's us split it up into the sum of two fractions with the same denominator.

so as an example, if:

a/x + b/x = (a + b) / x

Then the backwards is true:

(a + b) / x = a/x + b/x

So we can split our expression into the difference of two fractions with the same denominator:

a(b + 1) / (b + 1) - 2 / (b + 1)

And this is why I never simplified the a(b + 1) term. Now the (b + 1) terms cancel out in the first fraction, giving us:

a - 2 / (b + 1)

Which is what you said we should get and didn't know how to get there.

Hope this helped. If it did, please give best answer. Thanks.

[A[1+(1/B)]-(2/B)] / [1+(1/B)] 

becomes 

A-[2/(B+1)]

This is very simple. By fractional expansion we have...

[ A[1+(1/B)] / [1+(1/B)] ] - [ (2/B) / [1+(1/B)] ]

as similar terms cancel in the numerator and the denominator. Leaving just...

A - (2/B) / [1+(1/B)]

Multiply the second term by one, or B/B

A - (B/B) (2/B) / [1+(1/B)]

which yields the desired answer...

A - [2/(B + 1)]

The denominator cancels the multiplier of A in the numerator, so you’re left with

.. A - (2/B) / (1 +1/B)

.. = A -(2/B)/((B +1)/B) . . . . combine denominator terms

.. = A -2/(B +1) . . . . multiply by B/B