5. [12x/ (5xy^4)] - [9y^2/(15x^3y^2)]

6. [6/(x+2)] + 11/ (x-7)

- Jun 1st 2008, 03:44 PMgnarlycarly227Can somebody please show me how to solve these 2 problems step by step?
5. [12x/ (5xy^4)] - [9y^2/(15x^3y^2)]

6. [6/(x+2)] + 11/ (x-7) - Jun 1st 2008, 06:35 PMTheEmptySetQuote:

6. [6/(x+2)] + 11/ (x-7)

So now we build the fraction to had the LCD

$\displaystyle \frac{6}{(x+2)}\frac{(x-7)}{(x-7)}+\frac{11}{(x-7)}\frac{(x+2)}{(x+2)}=\frac{6x-42}{ (x-7)(x+2) }+\frac{11x+22}{(x-7)(x+2)}$

$\displaystyle =\frac{17x-20}{(x-7)(x+2)}$ - Jun 2nd 2008, 03:28 PMAlgebraAnswers#5 Answer
[12x/ (5xy^4)] - [9y^2/(15x^3y^2)]

LCD = 15 x^3 y^4

=[12x (3 x ^2) / 15 x^3 y^4] - [9y^2 (y^2) / 15 x^3 y ^4]

=[36 x^3 / 15x^3y^4] - [9y^4 / 15x^3y^4]

= 36x^3 - 9y^4 / 15x^3y^4 - Jun 2nd 2008, 06:37 PMTheEmptySet