simplify the question with the restriction
x^2+9y^2/x-3y + 6xy/3y-x
Assuming the usual carelessness with parentheses, we have
(x^2+9y^2)/(x-3y) + 6xy/(3y-x)
= (x^2-7xy+9y^2)/(x-3y)
= (x-3y)^2/(x-3y)
= x-3y
as long as x-3y ≠ 0
To simplify the expression (x^2 + 9y^2)/(x - 3y) + (6xy)/(3y - x) with the given restriction, we can start by factoring the numerator of the first fraction and the denominator of the second fraction.
First, let's factor the numerator of the first fraction, which is x^2 + 9y^2. This can be factored as (x + 3y)(x - 3y).
Next, we factor the denominator of the second fraction, which is 3y - x. This can be written as -(x - 3y).
Now, let's simplify the expression:
(x^2 + 9y^2)/(x - 3y) + (6xy)/(3y - x)
= [(x + 3y)(x - 3y)]/(x - 3y) + (6xy)/[-(x - 3y)] (substituting the factored forms)
= (x + 3y) + (-6xy)/(x - 3y) (canceling out the common factors)
= x + 3y - (6xy)/(x - 3y) (rearranging the terms)
So, the simplified expression with the given restriction is x + 3y - (6xy)/(x - 3y).