Is there a way of thinking when solving for the unknown variables using the partial derivatives from the Lagrange function?

Question

For example,

$L=5xy+8xz+3yz-λ(2xyz+1920)$

$L(x) = 5y+8z-2λyz=0$

$L(y) = 5x+3z-2λxz=0$

$L(z) = 8x+3y-2λxy=0$

$L(λ) = -2xyz+1920=0$

Where do I go from here?

Answer 1

$$ \begin{cases} -2 \lambda y z+5 y+8 z=0\\ -2 \lambda x z+5 x+3 z=0\\ -2 \lambda x y+8 x+3 y=0\\ 1920-2 x y z=0\\ \end{cases} $$ $$ \begin{cases} \lambda = \frac{5 y+8 z}{2 y z}\\ -2 \frac{5 y+8 z}{2 y z} x z+5 x+3 z=0\\ -2 \frac{5 y+8 z}{2 y z} x y+8 x+3 y=0\\ 1920-2 x y z=0\\ \end{cases} $$ $$ \begin{cases} \lambda = \frac{5 y+8 z}{2 y z}\\ -\frac{x (5 y+8 z)}{y}+5 x+3 z=0\to -z (8 x-3 y)=0 \\ -\frac{x (5 y+8 z)}{z}+8 x+3 y=0\to -y (5 x-3 z)=0\\ 1920-2 x y z=0\\ \end{cases} $$ $$ \begin{cases} \lambda = \frac{5 y+8 z}{2 y z}\\ y= \frac{8 x}{3}\\ z=\frac{5x}{3}\\ 1920-2 x \frac{8 x}{3} \frac{5x}{3}=0\to x^3=216\to x=6\\ \end{cases} $$ $$ \begin{cases} \lambda = \frac{1}{2 }\\ y= 16\\ z=10\\ x=6\\ \end{cases} $$