#include<bits/stdc++.h>
using namespace std;

const double eps=0.000000001;

double jz[100][100];
int n;

bool check0(double x){return fabs(x-0)<eps;}

void dbg(){
    for (int i=1;i<=n;i++){for (int j=1;j<=n+1;j++)cout << jz[i][j] << " ";cout << "\n";}
    cout << "\n";
}

signed main(){
    cin >> n;
    for (int i=1;i<=n;i++)for (int j=1;j<=n+1;j++)cin >> jz[i][j];
    int rk=0,i=1;
    for (int t=1;t<=n;t++){             //依次消xt
        //通过交换找到系数不为0的xi
        int j=i;
        while(j<=n&&check0(jz[j][t]))j++;
        if (j>n)continue;               //没有该主元直接跳过
        swap(jz[i],jz[j]);
        //化1
        double tmp=jz[i][t];
        for (int j=1;j<=n+1;j++){
            jz[i][j]/=tmp;
        }
        //dbg();
        //消掉别的行
        for (int j=1;j<=n;j++){         //被消xt的是第j行
            if (j==i)continue;
            double tmp=-jz[j][t];
            for (int k=1;k<=n+1;k++){     //当前处理第j行的xk
                jz[j][k]+=jz[i][k]*tmp;
            }
        }
        rk++;
        i++;
        //dbg();
    }
    //判断解的情况
    //先判断无解
    for (int i=1;i<=n;i++){
        bool ff=0;
        for (int j=1;j<=n;j++)if (!check0(jz[i][j]))ff=1;
        if (!ff){
            if (!check0(jz[i][n+1])){       
                cout << -1;
                return 0;
            }
        }
    }
    //是否存在无穷解
    if (rk<n){
        cout << 0;
        return 0;
    }
    for (int i=1;i<=n;i++){
        printf("x%d=%.2lf\n",i,jz[i][n+1]);
    }
	return 0;
}