# Consider the following system.\begin{cases}y=30-x\\2y=-x^{2}+16x-12\end{cases}

Consider the following system.
$$\begin{cases}y=30-x\\2y=-x^{2}+16x-12\end{cases}$$

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$$\displaystyle{y}={30}-{x}\rightarrow{\left({1}\right)}$$
$$\displaystyle{2}{y}=-{x}^{{{2}}}+{16}{x}-{12}\rightarrow{\left({2}\right)}$$
from 1 and 2 we have
$$\displaystyle{2}{\left({30}-{x}\right)}=-{x}^{{{2}}}+{16}{x}-{12}$$
$$\displaystyle{60}-{2}{x}=-{x}^{{{2}}}+{16}{x}-{12}$$
$$\displaystyle{x}^{{{2}}}-{16}{x}-{2}{x}+{12}+{60}={0}$$
$$\displaystyle{x}^{{{2}}}-{18}{x}+{72}={0}$$
$$\displaystyle{x}^{{{2}}}-{6}{x}-{12}{x}+{72}={0}$$
$$\displaystyle{x}{\left({x}-{6}\right)}-{12}{\left({x}-{6}\right)}={0}$$
$$\displaystyle{\left({x}-{6}\right)}{\left({x}-{12}\right)}={0}$$
$$\displaystyle\because{x}-{6}={0},{x}-{12}={0}$$
$$\displaystyle\because{x}={6},{x}={12}$$
when $$\displaystyle{x}={6},{y}={30}-{6}={24}{\left[{\mathfrak{{o}}}{m}{1}\right]}$$
when $$\displaystyle{x}={12},{y}={30}-{12}={18}{\left[{\mathfrak{{o}}}{m}{1}\right]}$$
Hence the requined solution
$$\displaystyle{\left({6},{24}\right)}$$ and $$\displaystyle{\left({12},{18}\right)}$$
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