Let's assume that the given numbers on the left of equals are coefficients of 5 variables: v, w, x, y, z, and the coefficients are generally represented by a, b, c, d, e, f. Let p represent the number on the right side of equals, so:
av+bw+cx+dy+ez+f=p. When a=b=c=d=e=0 then p=1, so f=1. Plugging in the coefficients for the other equations:
(1) v+3w+5x+4y+2z+1=56;
(2) v+2w+3x+2y+z+1=13, therefore (3) w+2x+2y+z=43 by subtraction;
(4) 3v+w+7x+2y+6z+1=106; therefore (5) 2v-w+4x+5z=93 (that is, (4)-(2));
(6) 4v+7w+8x+y+z+1=233.
We have 4 main equations but 5 variables so let's take one variable and treat it like a constant. Let that variable be v.
(1a) 3w+5x+4y+2z=55-v;
(2a) 2w+3x+2y+z=12-v;
(4a) w+7x+2y+6z=105-3v;
More to follow in due course...
(6)