#include <bits/stdc++.h>
using namespace std;
int main(){
double a,b,c,d,e,f;
while(cin>>a>>b>>c>>d>>e>>f){
if(a*e==d*b){
if(a*f==c*d)cout << "Too many\n";
else cout << "No answer\n";
}else{
double x=(c*e-b*f)/(a*e-b*d);
double y=(c*d-a*f)/(b*d-a*e);
cout << "x=" << fixed << setprecision(2) << x << endl;
cout << "y=" << fixed << setprecision(2) << y << endl;

}
}
}
/*求ｙ
adx+bdy=cd...1
adx+aey=af...2

2-1得(bd-ae)y=cd-af
y=(cd-af)/(bd-ae)

求ｘ
aex+bey=ce
bdx+bey=bf

同理，
x=(ce-bf)/(ae-bd)
*/