Solve for \(x\) in \(x+y=-6\).
Solve for \(x\).
\[x+y=-6\]
Subtract \(y\) from both sides.
\[x=-6-y\]
\[x=-6-y\]
Substitute \(x=-6-y\) into \({x}^{2}+{y}^{2}=216\).
Start with the original equation.
\[{x}^{2}+{y}^{2}=216\]
Let \(x=-6-y\).
\[{(-6-y)}^{2}+{y}^{2}=216\]
\[{(-6-y)}^{2}+{y}^{2}=216\]
Solve for \(y\) in \({(-6-y)}^{2}+{y}^{2}=216\).
Solve for \(y\).
\[{(-6-y)}^{2}+{y}^{2}=216\]
Expand.
\[{(-6)}^{2}-2\times -6y+{y}^{2}+{y}^{2}=216\]
Since the power of 2 is even, the result will be positive.
\[{6}^{2}-2\times -6y+{y}^{2}+{y}^{2}=216\]
Simplify \({6}^{2}\) to \(36\).
\[36-2\times -6y+{y}^{2}+{y}^{2}=216\]
Simplify \(2\times -6y\) to \(-12y\).
\[36-(-12y)+{y}^{2}+{y}^{2}=216\]
Remove parentheses.
\[36+12y+{y}^{2}+{y}^{2}=216\]
Simplify \(36+12y+{y}^{2}+{y}^{2}\) to \(36+12y+2{y}^{2}\).
\[36+12y+2{y}^{2}=216\]
Move all terms to one side.
\[36+12y+2{y}^{2}-216=0\]
Simplify \(36+12y+2{y}^{2}-216\) to \(-180+12y+2{y}^{2}\).
\[-180+12y+2{y}^{2}=0\]
Use the Quadratic Formula.
In general, given \(a{x}^{2}+bx+c=0\), there exists two solutions where:
\[x=\frac{-b+\sqrt{{b}^{2}-4ac}}{2a},\frac{-b-\sqrt{{b}^{2}-4ac}}{2a}\]
In this case, \(a=2\), \(b=12\) and \(c=-180\).
\[{y}^{}=\frac{-12+\sqrt{{12}^{2}-4\times 2\times -180}}{2\times 2},\frac{-12-\sqrt{{12}^{2}-4\times 2\times -180}}{2\times 2}\]
Simplify.
\[y=\frac{-12+12\sqrt{11}}{4},\frac{-12-12\sqrt{11}}{4}\]
\[y=\frac{-12+12\sqrt{11}}{4},\frac{-12-12\sqrt{11}}{4}\]
Simplify solutions.
\[y=-3(1-\sqrt{11}),-3(1+\sqrt{11})\]
\[y=-3(1-\sqrt{11}),-3(1+\sqrt{11})\]
Substitute \(y=-3(1-\sqrt{11}),-3(1+\sqrt{11})\) into \(x=-6-y\).
Start with the original equation.
\[x=-6-y\]
Let \(y=-3(1-\sqrt{11}),-3(1+\sqrt{11})\).
\[x=-6+3(1-\sqrt{11}),-6+3(1+\sqrt{11})\]
Simplify.
\[x=-3-3\sqrt{11},-3+3\sqrt{11}\]
\[x=-3-3\sqrt{11},-3+3\sqrt{11}\]
Therefore,
\[\begin{aligned}&x=-3-3\sqrt{11},-3+3\sqrt{11}\\&y=-3(1-\sqrt{11}),-3(1+\sqrt{11})\end{aligned}\]
x=-3-3*sqrt(11),-3+3*sqrt(11);y=-3*(1-sqrt(11)),-3*(1+sqrt(11))
