Question

Simplify and answer the boxes.

Answer 1

Answer:

$\huge\boxed{\dfrac{x^2+9^2}{x-3y}+\dfrac{6xy}{3y-x}=x-3y}$

Step-by-step explanation:

Domain:

$x-3y\neq0\Rightarrow x\neq3y$

$\dfrac{x^2+9y^2}{x-3y}+\dfrac{6xy}{3y-x}=\dfrac{x^2+9y^2}{x-3y}+\dfrac{6xy}{-(x-3y)}\\\\=\dfrac{x^2+9y^2}{x-3y}-\dfrac{6xy}{x-3y}=\dfrac{x^2+9y^2-6xy}{x-3y}\\\\=\dfrac{x^2-2(x)(3y)+(3y)^2}{3y-x}=\dfrac{(x-3y)^2}{3y-x}\\\\=\dfrac{\bigg[-1(3y-x)\bigg]^2}{3y-x}=\dfrac{(-1)^2(3y-x)^2}{3y-x}\\\\=\dfrac{1(x-3y)(x-3y)}{x-3y}=x-3y$

Used:

The distributive property: a(b + c) = ab + ac

(a - b)² = a² - 2ab + b²