Need help with adding rational expressions

[mcruz65]

3x + 3y + 3  / x^2 + 2xy + y^2 - 1
------------- ----------------------
9x             x^4 + x^2

3(x + y + 1) . x^4 + x^2
------------- ---------------------------
9x             x^2 + 2xy + y^2 -1

1(x + y + 1) . x^2(x^2 + 1)
------------- -------------------------------
3x             x(x + 2xy) + (y + 1) (y - 1)

x + y + 1 . ??????
---------- -----------------------
3x

 

[Denis]

x^2 + 2xy + y^2 - 1 = (x + y + 1)(x + y - 1) ; carry on...

 

[masters]

Hi mcruz65,

I'll pick it up after you inverted and use LaTex to make it more readible.

\(\displaystyle \frac{3(x+y+1)}{9x}\cdot \frac{x^4+x^2}{\underbrace{x^2+2xy+y^2}_{\text{perfect square trinomial}}-1}\)

\(\displaystyle =\frac{3(x+y+1)}{9x} \cdot \frac{x^2(x^2+1)}{\underbrace{(x+y)^2-1}_{\text{difference of squares}}}\)

\(\displaystyle =\frac{3(x+y+1)}{9x} \cdot \frac{x^2(x^2+1)}{(x+y+1)(x+y-1)}\)

Does that give you a little more to work with?

 

[mcruz65]

x + y + 1 . x^2(x^2+1)
---------- ----------------------
   3x      (x + y + 1)(x + y + 1)

    1     . x (x^2 + 1)
---------- ----------------------
    3       x + y + 1

x (x^2 + 1)
------------
3x + 3y + 3

Is this correct?

 

[mmm4444bot]

x + y + 1 . x^2(x^2+1)
---------- ----------------------
   3x      (x + y + 1)(x + y + 1) <-------- this should be -1

    1     . x (x^2 + 1)
---------- ----------------------
    3       x + y + 1

x (x^2 + 1)
------------
3x + 3y + 3 <------------ this should be -3

Is this correct?

If you're going to distribute the 3 on the bottom, then you could distribute the x on the top, too. Or, you could leave everything factored.

\(\displaystyle \frac{x^3 + x}{3x + 3y - 3}\)

OR

\(\displaystyle \frac{x(x^2 + 1)}{3(x + y - 1)}\)