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

zi2lalZ 2021-09-19 Answered

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

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Expert Answer

timbalemX
Answered 2021-09-20 Author has 10810 answers
\(\displaystyle{y}={30}-{x}\rightarrow{\left({1}\right)}\)
\(\displaystyle{2}{y}=-{x}^{{{2}}}+{16}{x}-{12}\rightarrow{\left({2}\right)}\)
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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