12. Solve the system of equations shown below algebraically: $$ \begin{array}{l}(x-3)^{2}+(y+2)^{2}=16 \ 2 x+2 y=10\end{array} $$

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Solve the system of equations $$ (x-3)^{2}+(y+2)^{2}=16;2x+2y=10 $$.
Solve the system of equations by following steps:
- step0: Solve using the substitution method:
  $$\left\{ \begin{array}{l}\left(x-3ight)^{2}+\left(y+2ight)^{2}=16\2x+2y=10\end{array}ight.$$
- step1: Solve the equation:
  $$\left\{ \begin{array}{l}\left(x-3ight)^{2}+\left(y+2ight)^{2}=16\x=5-y\end{array}ight.$$
- step2: Substitute the value of $$x:$$
  $$\left(5-y-3ight)^{2}+\left(y+2ight)^{2}=16$$
- step3: Simplify:
  $$8+2y^{2}=16$$
- step4: Move the constant to the right side:
  $$2y^{2}=16-8$$
- step5: Subtract the numbers:
  $$2y^{2}=8$$
- step6: Divide both sides:
  $$\frac{2y^{2}}{2}=\frac{8}{2}$$
- step7: Divide the numbers:
  $$y^{2}=4$$
- step8: Simplify the expression:
  $$y=\pm \sqrt{4}$$
- step9: Simplify:
  $$y=\pm 2$$
- step10: Separate into possible cases:
  $$y=2\cup y=-2$$
- step11: Rearrange the terms:
  $$\left\{ \begin{array}{l}x=5-y\y=2\end{array}ight.\cup \left\{ \begin{array}{l}x=5-y\y=-2\end{array}ight.$$
- step12: Calculate:
  $$\left\{ \begin{array}{l}x=3\y=2\end{array}ight.\cup \left\{ \begin{array}{l}x=7\y=-2\end{array}ight.$$
- step13: Check the solution:
  $$\left\{ \begin{array}{l}x=3\y=2\end{array}ight.\cup \left\{ \begin{array}{l}x=7\y=-2\end{array}ight.$$
- step14: Rewrite:
  $$\left(x,yight) = \left(3,2ight)\cup \left(x,yight) = \left(7,-2ight)$$
The system of equations has two solutions:

1. $$ (x, y) = (3, 2) $$
2. $$ (x, y) = (7, -2) $$
The solutions are $$ (3, 2) $$ and $$ (7, -2) $$.

Solve the system of equations shown below algebraically:

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