# Substract 9x^2y^2-3x^2-5 from the sum of -6x^2y^2-x^2-1 and 5x^2y^2+2x^2-1

Substract $$\displaystyle{9}{x}^{{2}}{y}^{{2}}-{3}{x}^{{2}}-{5}$$ from the sum of $$\displaystyle-{6}{x}^{{2}}{y}^{{2}}-{x}^{{2}}-{1}$$ and $$\displaystyle{5}{x}^{{2}}{y}^{{2}}+{2}{x}^{{2}}-{1}$$

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First, find the sum of the polynomials $$\displaystyle-{6}{x}^{{2}}{y}^{{2}}-{x}^{{2}}-{1}$$ and $$\displaystyle{5}{x}^{{2}}{y}^{{2}}+{2}{x}^{{2}}-{1}$$
Then,
$$\displaystyle{\left(-{6}{x}^{{2}}{y}^{{2}}-{x}^{{2}}-{1}\right)}+{\left({5}{x}^{{2}}{y}^{{2}}+{2}{x}^{{2}}-{1}\right)}=-{6}{x}^{{2}}{y}^{{2}}+{5}{x}^{{2}}{y}^{{2}}-{x}^{{2}}+{2}{x}^{{2}}-{1}-{1}$$
$$\displaystyle={\left(-{6}+{5}\right)}{x}^{{2}}{y}^{{2}}+{\left(-{1}+{2}\right)}{x}^{{2}}-{2}$$
$$\displaystyle=-{x}^{{2}}{y}^{{2}}+{x}^{{2}}-{2}$$
Therefore, the sum of the polynomials $$\displaystyle-{6}{x}^{{2}}{y}^{{2}}-{x}^{{2}}-{1}$$ and $$\displaystyle{5}{x}^{{2}}{y}^{{2}}+{2}{x}^{{2}}-{1}$$ is $$\displaystyle-{x}^{{2}}{y}^{{2}}+{x}^{{2}}-{2}$$
Now, subtract $$\displaystyle{9}{x}^{{2}}{y}^{{2}}-{3}{x}^{{2}}-{5}$$ from $$\displaystyle-{x}^{{2}}{y}^{{2}}+{x}^{{2}}-{2}$$
So,
$$\displaystyle{\left(-{x}^{{2}}{y}^{{2}}+{x}^{{2}}-{2}\right)}-{\left({9}{x}^{{2}}{y}^{{2}}-{3}{x}^{{2}}-{5}\right)}=-{x}^{{2}}{y}^{{2}}+{x}^{{2}}-{2}-{9}{x}^{{2}}{y}^{{2}}+{3}{x}^{{2}}+{5}$$
$$\displaystyle=-{x}^{{2}}{y}^{{2}}-{9}{x}^{{2}}{y}^{{2}}+{x}^{{2}}+{3}{x}^{{2}}-{2}+{5}$$
$$\displaystyle={\left(-{1}-{9}\right)}{x}^{{2}}{y}^{{2}}+{\left({1}+{3}\right)}{x}^{{2}}+{5}-{2}$$
$$\displaystyle=-{10}{x}^{{2}}{y}^{{2}}+{4}{x}^{{2}}+{3}$$
Hence, the required polynomial is $$\displaystyle-{10}{x}^{{2}}{y}^{{2}}+{4}{x}^{{2}}+{3}$$