Go: package main

import "fmt"

func main() {
	var resultX, resultY [100]float32
	var count int
	for {
		var a, b, c, d, e, f, x, y float32
		fmt.Scanf("%f %f %f %f %f %f", &a, &b, &c, &d, &e, &f)
		if a == 0 {
			break
		}
		//Simultaneous equation
		y = (((a * f) - (d * c)) / ((a * e) - (b * d)))
		x = ((c - (b * y)) / a)
		resultX[count] = x
		resultY[count] = y
		count++
	}
	for i := 0; i < count; i++ {
		fmt.Printf("%.3f %.3f\n", resultX[i], resultY[i])
	}
}

C++: #include <iostream>
using namespace std;

int main(){
	int a, b, c, d, e, f;

	while(cin >> a >> b >> c >> d >> e >> f){
		float x, y;
		x = 1.0*(b*f - c*e)/(b*d -a*e);
		y = 1.0*(c - a*x)/b;
		cout << x << " " << y << endl;
	}
	return 0;
}

C++: #include <iostream>
#include <cstdio>
using namespace std;
int main(void){
     
    double x, y;
    int a, b, p;
    int c, d, q;
 
    while (cin >> a >> b >> p >> c >> d >> q){
 
        x = (double)((p * d - b * q) / (a * d - b * c));
        y = (double)((a * q + p * c) / (a * d - b * c));
        printf("%.3f %.3f\n", x, y);
    }
}

C++: #include <iostream>
#include <stdio.h>
using namespace std;

int main(){
	double a, b, c, d, e, f;
	double x, y;

	while (cin >> a >> b >> c >> d >> e >> f){
		y = (f*a - c*d) / (e*a - b*d);
		x = (c - b*y) / a;
		printf("%.3lf %.3lf\n", x, y);
	}
}

C++: #include <iostream>
#include <stdio.h>
using namespace std;

int main()
{
  int a,b,c,d,e,f;
  while(cin >> a >> b >> c >> d >> e >> f){
    double x,y;
    x = (double)(c*e-f*b)/(a*e-d*b);
    y = (double)(c*d-f*a)/(b*d-e*a);
    printf("%.3f %.3f\n", x,y);
  }
  return 0;
}

C++: #include <bits/stdc++.h>

#define rep(i,n) for(long long int (i)=0;(i)<(int)(n);(i)++)
#define rrep(i,a,b) for(long long int i=(a);i<(b);i++)
#define rrrep(i,a,b) for(long long int i=(a);i>=(b);i--)
#define all(v) (v).begin(), (v).end()
#define pb(q) push_back(q)
#define Abs(a,b) max(a,b)-min(a,b)
#define YES(condition) if(condition){cout << "YES" << endl;}else{cout << "NO" << endl;}
#define Yes(condition) if(condition){cout << "Yes" << endl;}else{cout << "No" << endl;}
#define Cout(x) cout<<(x)<<endl
#define POSSIBLE(condition) if(condition){cout << "POSSIBLE" << endl;}else{cout << "IMPOSSIBLE" << endl;}
#define Possible(condition) if(condition){cout << "Possible" << endl;}else{cout << "Impossible" << endl;}
#define possible(condition) if(condition){cout << "possible" << endl;}else{cout << "impossible" << endl;}
#define Size(n) (n).size()

typedef long long ll;


using namespace std;

const int INF = 1e9,MOD = 1e9 + 7,ohara=1e6;
const ll LINF = 1e18;

/*---------------------------------------------------------------
long long int kaizyo(long long int hh){
    cmp=1;
    while(hh>1){
        cmp=(cmp*hh)%MOD;
        hh--;
    }
    return cmp;
}
long long int ruizyo(long long int aa, long long int bb){
    if(aa==0){
        return 1;
    }
    else if(aa%2==0){
        long long int tt=ruizyo(aa/2,bb);
        return (tt*tt)%MOD;
    }
    else{
        return (ruizyo(aa-1,bb)*bb)%MOD;
    }
}フェルマ－のア
---------------------------------------------------------------

while(x!=0){
            sum+=x%10;
          / x/=10;
        }
        各桁の和
---------------------------------------------------------------

pair<int,int> p[100000];
cin >> tmp;
p[i]=make_pair(tmp,i);
cout << p[i].second+1 << endl;//ペアの右側つまりiを出力

---------------------------------------------------------------

bool f[100001];//1000000以下の素数を調べよう！
rrep(i,2,100001){
       f[i]=false;
    }
rrep(i,2,100001){
        if(!f[i]){
            for(int j=i+i;j<=100000;j+=i){
                f[j]=true;
            }
        }
    }
    for(int i=3;i<=100000;i+=2){
        if(!f[i]){
            c[i]++;
        }
    }

---------------------------------------------------------------

long long gcd(long long aaa,long long bbb){
    if(bbb==0){
        return aaa;
    }
    return gcd(bbb,aaa%bbb);
}
 
long long lcm(long long aaa,long long bbb){
    long long g = gcd(aaa,bbb);
    return aaa/g * bbb;
}左から最大公約数と最小公倍数


---------------------------------------------------------------

long long int prime_cnt[10000];
       for(int i=2;i*i<=n;i++){
           while(n%i==0){
               n/=i;
               prime_cnt[i]+=1;
           }
           if(n>1){
               prime_cnt[n]+=1;
               break;
           }
      }ある数nを素因数分解しましょう

---------------------------------------------------------------
debug:
cout<<"i: "<<i<<" j: "<<j<<" cnt: "<<cnt<<"\n";

---------------------------------------------------------------*/

long long int n,cnt=0,ans=0,cmp,cmpp,m,h,w,x,y,sum=0,pos;
int dy[]={1,0,-1,0};
int dx[]={0,1,0,-1};
string alph("abcdefghijklmnopqrstuvwxyz"),s,co;
bool fl=true;
struct edge{int to,cost;};

//-------------------------↓↓↓↓↓↓------------------------

int main(void){
       cin.tie(0);
    ios::sync_with_stdio(false);

      double a,b,c,d,e,f;
      while(cin>>a>>b>>c>>d>>e>>f){
          double x,y;
          y=(a*f-d*c)/(a*e-b*d);
          x=(f-e*y)/d;
          if(x==-0)x=0;
          printf("%.15f %.15f\n",x,y);
      }
       
       
return 0;
}

C++: #include<bits/stdc++.h>
#include<stdio.h>
using namespace std;
int main()
{
    float a,b,c,d,e,f;
    float x=0,y=0;
    while(scanf("%f %f %f %f %f %f",&a,&b,&c,&d,&e,&f)!=EOF)
    {
        x=(c*e-b*f)/(a*e-b*d);
        y=(a*f-c*d)/(a*e-b*d);
        if(x==-0.000000)
            x=0.000000;
        if(y==-0.000000)
            y=0.000000;
        printf("%.3f %.3f\n",x,y);
    }
}

C++: #include <stdio.h>
#include <cmath>
using namespace std;

int main(void) {
  float a, b, c, d, e, f;
  while (scanf("%f %f %f %f %f %f", &a, &b, &c, &d, &e, &f)+1)  {
    float x = (c*e-b*f)/(a*e-b*d);
    float y = (c*d-a*f)/(b*d-a*e);
    if (abs(x) < -1.0e-4) x = 0.0;
    if (abs(y) < -1.0e-4) y = 0.0;
    printf("%.3f %.3f\n", x, y);
  }
  return 0;
}

C++: #include<stdio.h>
#include<iostream>
#include<iomanip>

#define REP(i, a) for(int i=0;i<(int)a;++i)
using namespace std;

void solver() {
    double a, b, c, d, e, f;
    while (cin >> a >> b >> c >> d >> e >> f) {
        auto y = (c * d - f * a) / (b * d - a * e);
        auto x = (c * e - f * b) / (a * e - b * d);
        cout << fixed;
        cout << setprecision(3);
        cout << x+0.0005 << " " << y+0.0005 << endl;
    }
}

int main() {
    solver();
    return 0;
}

C++: #include <iostream>
#include <iomanip>
#include <math.h>
using namespace std;

int main()
{
	double A, B, C, D, E, F, X, Y, G , H;

	while(cin >> A >> B >> C >> D >> E >> F)
	{
		G = A / D;
		D *= G;
		E *= G;
		F *= G;
		Y = (F-C) / (E-B);
		
		D /= G;
		E /= G;
		F /= G;
		H = B / E;
		D *= H;
		E *= H;
		F *= H;
		X = (F-C) / (D-A);

		X= floor(X*1000+0.5)/1000;
		Y= floor(Y*1000+0.5)/1000;

		if(Y==0){
			Y=0.000;
		}

		if(X==0){
			X=0.000;
		}

		cout<<fixed<<setprecision(3)<<X<<" "<<fixed<<setprecision(3)<<Y<<endl;
	}

	return 0;
}