Solve {(x-3)^2+(y+1)^2=5; x-3y=7}

Burhan Hopper 2021-05-16 Answered

Solve \(\begin{cases}(x-3)^2+(y+1)^2=5\\x-3y=7\end{cases}\)

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Expert Answer

liingliing8
Answered 2021-05-17 Author has 11570 answers

Let:
\(\displaystyle{\left({x}−{3}\right)}^{{2}}+{\left({y}+{1}\right)}^{{2}}={5}{\left({1}\right)}\)
\(x−3y=7\)
Use substitution. Solve for x using (2) to obtain (3):
\(x=3y+7(3)\)
Substitute (3) to (1) and solve for y:
\(\displaystyle{\left({3}{y}+{7}−{3}\right)}^{{2}}+{\left({y}+{1}\right)}^{{2}}={5}\)
\(\displaystyle{\left({3}{y}+{4}\right)}^{{2}}+{\left({y}+{1}\right)}^{{2}}={5}\)
\(\displaystyle{\left({9}{y}^{{2}}+{24}{y}+{16}\right)}+{\left({y}^{{2}}+{2}{y}+{1}\right)}={5}\)
\(\displaystyle{10}{y}^{{2}}+{26}{y}+{17}={5}\)
\(\displaystyle{10}{y}^{{2}}+{26}{y}+{12}={0}\)
Factor:
\(\displaystyle{2}{\left({5}{y}^{{2}}+{13}{y}+{6}\right)}={0}\)
\(\displaystyle{2}{\left({y}+{2}\right)}{\left({5}{y}+{3}\right)}={0}\)
By zero product property,
\(y=−2,−35\)
Solve for the corresponding \(xx\) values using (3).
When \(y=−2y=−2,\)
\(x=3(−2)+7\)
\(x=1\)
When \(y=−\frac{3}{5}\),
\(x=3(−\frac{3}{5})+7\)
\(x=\frac{26}{5}\)
So, the solutions are:
\((1,−2)\) and \((\frac{26}{5},−\frac{3}{5})\)

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Answered 2021-08-11 Author has 1944 answers

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