Math Problem Solver

The system of equations has $2$ solutions.

$$x = 6 + \sqrt{47}, y = -6 + \sqrt{47}$$$$x = 6 - \sqrt{47}, y = - \sqrt{47} - 6$$

By substituting these values into the expression, we find that: $$\begin{aligned}x^{2} + y^{2} &= 166\end{aligned}$$

Solve $x - y = 12$ for $x$. $$x = y + 12$$Substitute $y + 12$ for $x$ in $x y = 11$ and simplify. $$\begin{aligned}x y &= 11 \\ \left(y + 12\right) y &= 11 \\ y^{2} + 12 y &= 11 \\y^{2} + 12 y - 11 &= 0 \\ y &= \frac{-(12) \pm \sqrt{(12)^{2} - 4(1)(-11)}}{2(12)} \\ y = -6 + \sqrt{47}&, y = - \sqrt{47} - 6\end{aligned}$$Substitute $-6 + \sqrt{47}$ into $x - y = 12$ to solve for $x$. $$\begin{aligned}x - \sqrt{47} + 6 &= 12\\x + \left(6 - \sqrt{47}\right) &= 12\\x &= 6 + \sqrt{47}\end{aligned}$$This yields the following solution. $$\begin{aligned}x = 6 + \sqrt{47},\,y = -6 + \sqrt{47}\end{aligned}$$Substitute $- \sqrt{47} - 6$ into $x - y = 12$ to solve for $x$. $$\begin{aligned}x + 6 + \sqrt{47} &= 12\\x + \left(6 + \sqrt{47}\right) &= 12\\x &= 6 - \sqrt{47}\end{aligned}$$This yields the following solution. $$\begin{aligned}x = 6 - \sqrt{47},\,y = - \sqrt{47} - 6\end{aligned}$$

By substituting these values into the expression, we find that: $$\begin{aligned}x^{2} + y^{2} &= 166\end{aligned}$$