a+2b+3c+4d+5e=41
2a+3b+4c+5d+e=15
3a+4b+5c+d+2e=34
4a+5b+c+2d+3e=63
5a+b+2c+3d+4e=57
Thanks a lot :)
No solution I might be wrong can someone confirm?
Use eliminations ans substitutions to get:
a = 6, b = 4, c = - 3, d = - 1, e = 8
This solution balances the 5 equations.