Find all solutions of the system of equations.\begin{cases}x-2y=2\\y^{2}

Brittney Lord 2021-09-29 Answered

Find all solutions of the system of equations.
\(\begin{cases}x-2y=2\\y^{2}-x^{2}=2x+4\end{cases}\)

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Expert Answer

Gennenzip
Answered 2021-09-30 Author has 11665 answers

Step 1
To find:
The solution of the system of equation.
Given:
The system of equation \(x−2y=2\) and \(\displaystyle{y}^{{{2}}}−{x}^{{{2}}}={2}{x}+{4}\).
Calculation:
Simplify the equation \(x−2y=2\) as follows:
\(x−2y=2\)
\(x=2+2y\)
Substitute \(x=2+2y\) in \(\displaystyle{y}^{{{2}}}-{x}^{{{2}}}={2}{x}+{4}\).
\(\displaystyle{y}^{{{2}}}-{\left({2}{y}+{2}\right)}^{{{2}}}={2}{\left({2}{y}+{2}\right)}+{4}\)
\(\displaystyle{y}^{{{2}}}-{4}{y}^{{{2}}}-{4}-{8}{y}={4}{y}+{4}+{4}\)
\(\displaystyle-{3}{y}^{{{2}}}-{4}-{8}-{8}{y}-{4}{y}={0}\)
\(\displaystyle-{3}{y}^{{{2}}}-{12}-{12}{y}={0}\)
Further simplify as follows:
\(\displaystyle{y}^{{{2}}}+{4}{y}+{4}={0}\)
\(\displaystyle{\left({y}+{2}\right)}^{{{2}}}={0}\)
\(y=-2\)
Sbstitute \(y=-2\) in \(x=2+2y\).
\(x=2+2(-2)\)
\(=2-4\)
\(=-2\)
Step 2
Thus, the solution of the system of equation \(x=−2\) and \(y=−2.\)

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