z=2 x² + 6x²+3(x + y)² + 2 , where x=2, y =-2
You asked:
Solve the system of equations:\[\begin{aligned}z &= 2 \cdot {x}^{2} + 6 \cdot {x}^{2} + 3 \cdot {\left( x + y \right)}^{2} + 2\\x &= 2\\y &= -2\end{aligned}\]
MathBot Answer:
The system of equations has one solution.
\[x = 2, y = -2, z = 34\]