# Math Help - Can somebody please show me how to solve these 2 problems step by step?

1. ## Can 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)

2. 6. [6/(x+2)] + 11/ (x-7)
We find the LCD of the two fractions and that is (x+2)(x-7)

So now we build the fraction to had the LCD

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

$=\frac{17x-20}{(x-7)(x+2)}$

[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

[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
This can be simplified to

$\frac{36x^3-9y^4}{15x^3y^4}=\frac{9(4x^3-y^4)}{15x^3y^4}=\frac{3(4x^3-y^4)}{5x^3y^4}$