200z=(x2x^{2}/2)+(y2y^{2}/24)

20z=x

4z=y/12

asked by guest
on Nov 24, 2024 at 9:05 pm



You asked:

Solve the system of equations:200z=x22+y22420z=x4z=y12\begin{aligned}200 z &= \frac{{x}^{2}}{2} + \frac{{y}^{2}}{24}\\20 z &= x\\4 z &= \frac{y}{12}\end{aligned}

MathBot Answer:

The system of equations has 22 solutions.

x=0,y=0,z=0x = 0, y = 0, z = 0x=50037,y=120037,z=2537x = \frac{500}{37}, y = \frac{1200}{37}, z = \frac{25}{37}