From the wxMaxima issue tracker #690.
solve can't solve a system of two quadratic equations with no free parameters. This example should have two real solutions:
(%i1) solve([x^2+y^2=1, (x-2)^2+(y-3)^2=9],[x,y]);
(%o1) []
although solve can solve some examples like that:
(%i26) solve([x^2+y^2=1, x^2+(y-3)^2=9],[x,y]);
sqrt(35) 1 sqrt(35) 1
(%o26) [[x = - --------, y = -], [x = --------, y = -]]
6 6 6 6
(%i27) solve([x^2+y^2=1, (x-1)^2+(y-3)^2=9],[x,y]);
4 3
(%o27) [[x = - -, y = -], [x = 1, y = 0]]
5 5
For the record, Solver (loaded by load(solver);) finds a solution with x in terms of y and to_poly_solve can then find y:
(%i1) load (solver) $
(%i2) Solver ([x^2+y^2=1, (x-2)^2+(y-3)^2=9], [x,y]);
2 2
(%o2) [[x = - sqrt(1 - y ), [4 sqrt(1 - y ) - 6 y + 5]],
2 2
[x = sqrt(1 - y ), [(- 4 sqrt(1 - y )) - 6 y + 5]]]
(%i3) load (to_poly_solve) $
(%i4) map (lambda ([e], to_poly_solve (e, y)), map (second, %o2));
5/2 5/2
2 3 + 45 2 3 - 45
(%o4) [%union([y = -----------]), %union([y = - -----------])]
78 78
Duplicate of https://sourceforge.net/p/maxima/bugs/3208/
(%i3) eq1:x^2+y^2=1$
eq2:(x-2)^2+(y-3)^2=9$
solve([eq1,eq2-eq1],[x,y]);
(%o3)
[[x=-(3^(5/2)-10)/26,y=(23^(5/2)+45)/78],[x=(3^(5/2)+10)/26,y=-(23^(5/2)-45)/78]]
best
Aleksas D
2016-09-22 5:00 GMT+03:00 David Billinghurst billingd@users.sf.net:
Related
Bugs:
#3150Fixed by commit [2986fc] See [bugs:#3208] for details
Related
Bugs:
#3208Commit: [2986fc]
Last edit: David Billinghurst 2016-10-02