\(\displaystyle{2}{y}=-{x}^{{{2}}}+{16}{x}-{12}\rightarrow{\left({2}\right)}\)

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 \(\displaystyle{\left({12},{18}\right)}\)