# Solve [(9x^2 - 9y^2)/(6x^2y^2)]/[(3x + 3y)/(12x^2y^5)] .

hala718 | Certified Educator

(9x^2-9y^2)/(6x^2 y^2) / (3x+3y)/(12x^2 y^5)

= (9x^2 -9y^2)*(12x^2 y^5) / (6x^2 y^2)(3x+3y)

= 12 (3x-3y)(3x+3y)*(x^2 y^5) / 6(x^2 y^2)(3x+3y)

Now reduce similar:

= 2(3x-3y)*y^3

= 6(x-y) y^3

To solve the functions:

6(x-y)y^3=0

x=y    or  y=0

neela | Student

To solve

[(9x^2 - 9y^2)/(6x^2y^2)]/[(3x + 3y)/(12x^2y^5)] .

Solution:

We know (a/b)/ (c/d) = ad/(bc)

Rewriting  in numerator and denominator form we get: (9x^2-9y^2)(12x^2y^5)/ {(6x^2y^2)(3x+3y)}........(1)

Numerator = (3x+3y)(3x-3y)(1`2x^2y^5), as (a^2-b^2= (a+b)(a-b), wehre a= 3x and b= 3y and a^2-b^2 = 9x^2-9y^2.

Denominator =(3x+3y)(6x^2y^2)

Therefore reducing the expresion's ( at (1) ) numerator and denominator  (3x+3y)(6x^2y^2), the HCF of numerator and denominators:

(3x-3y)(2y^3) = 6(x-y)y^3

giorgiana1976 | Student

We'll calculate the square diffrenece:

9x^2-9y^2=(3x-3y)(3x+3y)

We'll write also 12x^2y^5=(6x*2)^2y^5=[(6x)^2y^5]*[(2)^2y^5]

We'll substitute 9x^2-9y^2 by (3x-3y)(3x+3y).

According to the rule,the division between 2 ratios is the product between the ratio from numerator and second back ratio from denominator.

The expression will become:

[(3x-3y)(3x+3y)]/[(6x)^2y^2] * {[(3x+3y)/[(6x)^2y^5]*[(2)^2y^5]}

We'll reduce similar terms and we'll get:

[(3x-3y)]*[(6x)^2y^3]*[(2)^2y^5]=[(3x-3y)]*[12x^2y^3]*(2)^4y