# Simplify [(6x^2*y^2)/(9x^2-9y^2)]/[(18x^2*y^3/(3x-3y)]

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To simplify

[(6x^2*y^2)/(9x^2-9y^2)]/[(18x^2*y^3/(3x-3y)]

Solution:

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

3x-3y = 3(x-y).

Therefore the given expression could re writtens as:

{6x^2y^2/9(x-y)(x+y)}/{18x^2y^3/(3(x-y)}

= (6x^2y^2)(3)(x-y)/ {9(x-y)(x+y)(18)(x^2y^3)}

Cancell the common factors . Or divide both numerator and denominators by 18x^2y^2(x-y).

= 1/{9y(x+y)} is the siplified form of the given expression

Here we have to simplify the expression

[(6x^2*y^2)/(9x^2-9y^2)]/[(18x^2*y^3)/(3x-3y)]

Now we know that a^2 - b^2 = ( a- b)(a+b)

[(6x^2*y^2)/(9x^2-9y^2)]/[(18x^2*y^3)/(3x-3y)]

arranging the denominators to simpplify the expression

=> [(6x^2*y^2)(3x-3y)] / [(9x^2-9y^2)(18x^2*y^3)]

cancelling 18x^2*y^2

=> (x-y) / (9x^2-9y^2)*y

=> (x-y) / 9y*(x^2 - y^2)

converting (x^2 - y^2)

=> (x-y) / 9y*(x-y)(x+y)

cancelling (x-y)

=> 1 / 9y*(x+y)

**Therefore the given expression simplified is 1/ 9y*(x+y)**

We notice that we have to calculate a division between 2 quotients. We'll have to multiply the first ratio with the inversed second ratio, where the numerator becomes denominator and denominator becomes numerator!

We'll re-write the denominator, that is a difference of squares, as a product:

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

The expression will become:

[(6x^2*y^2)/(3x-3y)(3x+3y)]*[(3x-3y)/(18x^2*y^3)]

We'll write the denominator 18x^2*y^3=3*6x^2*y^2*y

We can now simplify like terms, which are (3x-3y),and [(6x)^2y^2].

The result will be:

[1/(3x+3y)]*(1/3y)

**[(6x^2*y^2)/(9x^2-9y^2)]/[(18x^2*y^3/(3x-3y)]=1/9y(x+y)**