y,z=var('y,z'); solve(6*x + 10*y + 15*z ==1,x,y,z) gives
([{x: -5/3*y - 5/2*z + 1/6}], [1])while solve(x+y==3,x,y) gives ([x == -y + 3], [1]) My questions are: 1) Why the notation are different in the 2 and 3-variable case? One gives x: and the other x== 2) What the [1] in both cases stand for ? 3) In order to get the parameters appear in the solution, one need to add a redundant equation. Shall we improve on that? e.g. solve([x+y==3,2*x+2*y==6],x,y) [[x == -r1 + 3, y == r1]] -- To post to this group, send an email to sage-devel@googlegroups.com To unsubscribe from this group, send an email to sage-devel+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/sage-devel URL: http://www.sagemath.org
