Question

(-x^2-4x-5)+ (-x^2+8x+8)

Answer 1

Answer:

[\tex]\begin{lgathered}\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\end{lgathered}}[\tex]

Answer 2

Answer:

$\huge\boxed{-2x^2+4x+3}$

Step-by-step explanation:

$(-x^2-4x-5)+ (-x^2+8x+8)\\\\=-x^2-4x-5-x^2+8x+8\qquad|\text{combine like terms}\\\\=(-x^2-x^2)+(-4x+8x)+(-5+8)\\\\=-2x^2+4x+3$