//lg 2455
#include<bits/stdc++.h>
using namespace std;

const double eps = 0.000001;
const int N = 105;

double a[N][N];
int n;
int nowline=1;//存储当前行

void GaussJordan()
{
	for(int i=1;i<=n;i++) //枚举列 若方程有唯一解 则与枚举行列等效
	{
		int mxline=nowline;
		for(int k=i;k<=n;k++) if(fabs(a[k][i])>fabs(a[mxline][i])) mxline=k;
		if(fabs(a[mxline][i])<eps) continue;
		if(mxline!=nowline) swap(a[mxline],a[nowline]);
		for(int k=1;k<=n;k++)
		{
			if(k==nowline) continue;
			double t=a[k][i]/a[nowline][i];
			for(int j=i;j<=n+1;j++) a[k][j]-=t*a[nowline][j];
		}
		nowline++;
	}
	if(nowline==n+1)
	{
		for(int i=1;i<=n;i++) printf("x%d=%.2lf\n",i,a[i][n+1]/a[i][i]);
		return ;
	}
	while(nowline<n+1)
	{
		if(fabs(a[nowline][n+1])>eps) 
		{
			cout<<-1;
			return ;
		}
		nowline++;
	}
	cout<<0;
	return ;
}

int main()
{
	cin>>n;
	for(int i=1;i<=n;i++)
	{
		for(int j=1;j<=n+1;j++)
		{
			cin>>a[i][j];
		}
	}
	GaussJordan();
	return 0;
}