#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef long double LD;
#define int LL
#define double LD
typedef pair<int, int> PII;
const int N = 1e5 + 5;
const int INF = 0x3f3f3f3f;
const int inf = 0x3f3f3f3f3f3f3f3f;
const double PI = acosl(-1.0);
const double eps = 1e-8;
const int MOD = 1e9 + 7;

int n;
double a[110][110];
int gauss()
{
    int r, c;//行，列
    for (c = 1, r = 1; c <= n; c++) {
        int t = r;
        for (int i = r; i <= n; i++) {
            if (fabs(a[i][c]) > fabs(a[t][c])) t = i;
        }
        if (fabs(a[t][c]) < eps) continue;//用于下面判断解的情况
        for (int i = c; i <= n + 1; i++) swap(a[t][i], a[r][i]);
        for (int i = n + 1; i >= c; i--) a[r][i] /= a[r][c];
        for (int i = r + 1; i <= n; i++) {
            if (fabs(a[i][c]) > eps) {
                for (int j = n + 1; j >= c; j--) a[i][j] -= a[r][j] * a[i][c];//目的为了把第r的第c个数变为0
            }
        }
        r++;
    }
    if (r <= n) {
        for (int i = r; i <= n; i++) {
            if (fabs(a[i][n + 1]) > eps) return 2;//无解，出现0 = b情况
        }
        return 1;//无穷多解
    }
    for (int i = n; i >= 1; i--) {
        for (int j = i + 1; j <= n; j++) a[i][n + 1] -= a[i][j] * a[j][n + 1];
    }
    return 0;//有单组解
}
/*
for (int i = 1; i <= n; i++) {
		printf("%.2lf\n", a[i][n + 1]);
	}
*/
int p[10], q[10], r[10], s[10];
void solve()
{
    for (int i = 1; i <= 3; i++) {
        cin >> p[i] >> q[i] >> r[i] >> s[i];
    }
    p[2] += p[3];
    q[2] += q[3];
    r[2] += r[3];
    s[2] += s[3];
    cin >> p[0] >> q[0];
    n = 2;
    a[1][1] = p[1], a[1][2] = p[2], a[1][3] = p[0];
    a[2][1] = q[1], a[2][2] = q[2], a[2][3] = q[0];
    gauss();
    double ans1, ans2;
    ans1 = r[1] * a[1][3] + r[2] * a[2][3];
    ans2 = s[1] * a[1][3] + s[2] * a[2][3];
    cout << ans1 << " " << ans2 << "\n";
}

signed main()
{
    ios::sync_with_stdio(false), cin.tie(0);
    cout << fixed << setprecision(15);
    int _;
    cin >> _;
//    _ = 1;
    while (_--) {
        solve();
    }
    return 0;
}