100-2(x-y)-300( π\pi )(x-z)+100( π\pi )2y+2z(100( π\pi )) = 0

-300( π\pi )(z-x) -400( π\pi )(z-y)-5z-200( π\pi )(y)-400( π\pi )(z)+200( π\pi)(x) = 0

-2(y-x)-10y-400( π\pi )(y-z)+200( π\pi )(z)-200( π\pi )(x) = 0

asked by guest
on Mar 30, 2025 at 11:05 pm



You asked:

Solve the system of equations:100−2(x−y)−300π(x−z)+100π⋅2y+2⋅z(100π)=0−(300π(z−x))−400π(z−y)−5z−200πy−400πz+200πx=0−(2(y−x))−10y−400π(y−z)+200πz−200πx=0\begin{aligned}100 - 2 \left(x - y\right) - 300 \pi \left(x - z\right) + 100 \pi \cdot 2 y + 2 \cdot z\left( 100 \pi \right) &= 0\\-\left( 300 \pi \left(z - x\right) \right) - 400 \pi \left(z - y\right) - 5 z - 200 \pi y - 400 \pi z + 200 \pi x &= 0\\-\left( 2 \left(y - x\right) \right) - 10 y - 400 \pi \left(y - z\right) + 200 \pi z - 200 \pi x &= 0\end{aligned}

MathBot Answer:

The system of equations has one solution.

x=20⋅(3+760π+16000π2)(1+40π)(1+400π),y=10⋅(1+120π+8000π2)(1+40π)(1+400π),z=6400π(1+25π)(1+40π)(1+400π)x = \frac{20 \cdot \left(3 + 760 \pi + 16000 \pi^{2}\right)}{\left(1 + 40 \pi\right) \left(1 + 400 \pi\right)}, y = \frac{10 \cdot \left(1 + 120 \pi + 8000 \pi^{2}\right)}{\left(1 + 40 \pi\right) \left(1 + 400 \pi\right)}, z = \frac{6400 \pi \left(1 + 25 \pi\right)}{\left(1 + 40 \pi\right) \left(1 + 400 \pi\right)}

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