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KiRa [710]
3 years ago
6

Simplify and answer the boxes.

Mathematics
1 answer:
s2008m [1.1K]3 years ago
6 0

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

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(Picture attached)

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<h3>Learn more</h3>

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brainly.com/question/9909671

Keywords: linear inequality,graph

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