Rewritten these are:
8xy-3z=1159,
5x+5y+4z=100,
3xyz=-2145, so xyz=-715.
4xyz=-2860.
8xyz-3z²=1159z,
so 3z²+1159z-8xyz=0=3z²+1159z+5720.
Therefore (3z+1144)(z+5)=0, and z=-1144/3 or -5.
And xy=-715/z=143 or 15/8.
5x²+5xy+4xz=100x.
Let z=-5: 5x²+5×143-20x=100x.
So we have:
5x²-120x+5×143=0=x²-24x+143=(x-11)(x-13).
So for z=-5, x=11 or 13. Now, 5x+5y+4z=100, so:
5y=100-5x-4z and 5y=100-55+20=60, making y=12; or:
5y=100-65+20, making y=11.
This gives us (x,y,z)=(11,12,-5) or (13,11,-5).
But we need to test these in the original equations. Only (13,11,-5) satisfies the equations so is a solution.