O15-C908
> restart:
> c1:=x^2+x*y-2*y^2-4;c2:=x*exp(x)+y*exp(y);
> 

                             2            2
                      c1 := x  + x y - 2 y  - 4


                      c2 := x exp(x) + y exp(y)

> with(plots):implicitplot({c1,c2},x=-4..4,y=-4..4);

> implicitplot({c1,c2},x=-2.4..-2,y=0..0.5);

> # Il semble que (a,b)=(-2.13,0.21) à peu près. On précise par dichotomie :
> eps:=.001:ag:=-2.14;ad:=-2.12;bg:=solve(subs(x=ag,c1),y)[2];bd:=solve(subs(x=ad,c1),y)[2];
> while ad-ag>eps do
>                 am:=(ag+ad)/2;bm:=solve(subs(x=am,c1),y)[2];
>                 if evalf(subs({x=am,y=bm},c2)*subs({x=ag,y=bg},c2))<0 
>                                          then ag:=am;bg:=bm
>                                          else ad:=am;bd:=bm
>                 fi;
> od;

                             ag := -2.14


                             ad := -2.12


                          bg := .2239631085


                          bd := .1967048920


                          am := -2.130000000


                          bm := .2104712309


                          am := -2.135000000


                          bm := .2172503745


                          am := -2.137500000


                          bm := .2206149109


                          am := -2.138750000


                          bm := .2222910361


                          am := -2.139375000


                          bm := .2231275769

> 
> with(linalg):j:=matrix(2,2,[diff(c1,x),diff(c1,y),diff(c2,x),diff(c2,y)]);
Warning, new definition for norm
Warning, new definition for trace

                 [     2 x + y              x - 4 y     ]
            j := [                                      ]
                 [exp(x) + x exp(x)    exp(y) + y exp(y)]

> 
> f:=matrix(2,1,[c1,c2]);

                           [ 2            2    ]
                      f := [x  + x y - 2 y  - 4]
                           [                   ]
                           [x exp(x) + y exp(y)]

> x[0]:=matrix(2,1,[-2.1,0]);

                                    [-2.1]
                            x[0] := [    ]
                                    [ 0  ]

> for n from 1 to 5 do xx:=x[n-1][1,1];yy:=x[n-1][2,1];a:=subs({x=xx,y=yy},eval(j));x[n]:=evalm(x[n-1]-a^(-1)&*subs({x=xx,y=yy},eval(f))) od:
> seq(evalm(x[n]),n=1..5);

  [-2.129006579]  [-2.127304397]  [-2.126932765]  [-2.126932304]
  [            ], [            ], [            ], [            ],
  [.2532512530 ]  [.2081865564 ]  [.2062814385 ]  [.2062781556 ]

        [-2.126932305]
        [            ]
        [.2062781556 ]

> # Il apparait que la suite (X_n) semble converger vers le point fixe L : L=L-(df_L)^(-1)(f(L)) ie f(L)=0 donc L est le point d'intersection.