1. ## Some expressions

1. Simplify the expression
(2a^-2-2b^-2)/(4a^-1+4b^-1)
I know it goes to this
(2/a^2-2/b^2)/(4/a+4/b)
I eventually get it down to b-a/2ab
But I guess that's incorrect? I think it has to do with how I'm making the denominators equal.

(4x/x-1)+(4/1-x^2)-(2/x+1)
don't know how to find the lcd?
is the lcd (1+x)(1-x)(x-1)?

2. Hello, bilbobaggins!

2. Add/subtract: .$\displaystyle \frac{4x}{x-1} + \frac{4}{1-x^2} - \frac{2}{x+1}$
Factor a "minus" out of the middle denominator . . .

We have: .$\displaystyle \frac{4x}{x-1} - \frac{4}{x^2-1} - \frac{2}{x+1}$

Factor: .$\displaystyle \frac{4x}{x-1} - \frac{4}{(x-1)(x+1)} - \frac{2}{x+1}\quad\hdots\;\;\text{LCD} \:=\: (x-1)(x+1)$

Convert: .$\displaystyle \frac{4x}{x-1}\cdot{\color{blue}\frac{x+1}{x+1}} - \frac{4}{(x-1)(x+1)} - \frac{2}{x+1}\cdot{\color{blue}\frac{x-1}{x-1}}$

. . $\displaystyle = \;\frac{4x(x+1) - 4 - 2(x-1)}{(x-1)(x+1)} \;=\;\frac{4x^2+4x-4-2x+2}{(x-1)(x+1)} \;=\;\frac{4x^2+2x-2}{(x-1)(x+1)}$

Factor: .$\displaystyle \frac{2(2x^2+x-2)}{(x-1)(x+1)} \;=\;\frac{2(2x-1)(x+1)}{(x-1)(x+1)} \;=\; \frac{2(2x-1)}{x-1}$

3. Originally Posted by bilbobaggins
1. Simplify the expression
(2a^-2-2b^-2)/(4a^-1+4b^-1)
I know it goes to this
(2/a^2-2/b^2)/(4/a+4/b)
I eventually get it down to b-a/2ab
...
to #1.

$\displaystyle \left(2a^{-2}-2b^{-2}\right)\left(4a^{-1}+4b^{-1}\right)=8a^{-3}+8a^{-2}b^{-1} - 8a^{-1}b^{-2}-8b^{-3}$

The common denominator is $\displaystyle a^3b^3$. You'll get:

$\displaystyle \dfrac{8\left(-a^3+ab^2 -a^2b+b^3 \right)}{a^3b^3}=\dfrac{8\left(a(b^2-a^2)+b(b^2-a^2) \right)}{a^3b^3}=\dfrac{8(a-b)(a+b)^2}{a^3b^3}$

which you can't simplify any more.

4. Originally Posted by Soroban
Hello, bilbobaggins!

Factor a "minus" out of the middle denominator . . .

We have: .$\displaystyle \frac{4x}{x-1} - \frac{4}{x^2-1} - \frac{2}{x+1}$

Factor: .$\displaystyle \frac{4x}{x-1} - \frac{4}{(x-1)(x+1)} - \frac{2}{x+1}\quad\hdots\;\;\text{LCD} \:=\: (x-1)(x+1)$

Convert: .$\displaystyle \frac{4x}{x-1}\cdot{\color{blue}\frac{x+1}{x+1}} - \frac{4}{(x-1)(x+1)} - \frac{2}{x+1}\cdot{\color{blue}\frac{x-1}{x-1}}$

. . $\displaystyle = \;\frac{4x(x+1) - 4 - 2(x-1)}{(x-1)(x+1)} \;=\;\frac{4x^2+4x-4-2x+2}{(x-1)(x+1)} \;=\;\frac{4x^2+2x-2}{(x-1)(x+1)}$

Factor: .$\displaystyle \frac{2(2x^2+x-2)}{(x-1)(x+1)} \;=\;\frac{2(2x-1)(x+1)}{(x-1)(x+1)} \;=\; \frac{2(2x-1)}{x-1}$

I'm guessing the 2 right here 2(2x^2+x-2) was a typo?

Originally Posted by earboth
to #1.

$\displaystyle \left(2a^{-2}-2b^{-2}\right)\left(4a^{-1}+4b^{-1}\right)=8a^{-3}+8a^{-2}b^{-1} - 8a^{-1}b^{-2}-8b^{-3}$

The common denominator is $\displaystyle a^3b^3$. You'll get:

$\displaystyle \dfrac{8\left(-a^3+ab^2 -a^2b+b^3 \right)}{a^3b^3}=\dfrac{8\left(a(b^2-a^2)+b(b^2-a^2) \right)}{a^3b^3}=\dfrac{8(a-b)(a+b)^2}{a^3b^3}$

which you can't simplify any more.
Wait, what? Did you distribute the numbers? Like multiplication? I know with division of fractions you're supposed to flip then multiply. It doesn't appear you did that. Could you further explain?

Thanks for the help both of you.

5. Originally Posted by bilbobaggins
...

Wait, what? Did you distribute the numbers? Like multiplication? I know with division of fractions you're supposed to flip then multiply. It doesn't appear you did that. Could you further explain?

Thanks for the help both of you.
1. You have a product of sums therefore you have to use indeed the law of distribution.

2. The summands contains powers therefore you have to use the rules of calculating powers.

3. At this point I used the rules of adding fractions. I didn't write the fractions because I was pretty sure that you see that $\displaystyle 8a^{-2}b^{-1}=8\cdot \dfrac1{a^2} \cdot \dfrac1{b}$

4. The followings steps are nothing but factor out common factors from partial sums.