Solve the system by back substitution. \[ \begin{array}{r}2 x+4 y+z+3 w= \\ \qquad \begin{array}{l} x+6 z+2 w= \\ 4 z+4 w= \\ 3 w= \\ y\end{array} \\ \text { The solution is }(\square, \square, \square, \square) . ~(T y p e ~ i n t e g e r s ~ o r ~ s i m p l i f i e d ~ f r a c t i o n s .) ~\end{array} \]

To solve the system using back substitution, we first need to rewrite the equations clearly. 1. The equations are: \[ 3w = y \quad (1) \] \[ 4z + 4w = 0 \quad (2) \] \[ x + 6z + 2w = 0 \quad (3) \] \[ 2x + 4y + z + 3w = 0 \quad (4) \] Now, let’s work through them step-by-step. **From equation (1):** \[ y = 3w \] **From equation (2):** \[ 4z + 4w = 0 \implies z + w = 0 \implies z = -w \] **Substituting \(z = -w\) into equation (3):** \[ x + 6(-w) + 2w = 0 \implies x - 6w + 2w = 0 \implies x - 4w = 0 \implies x = 4w \] Now, we've expressed \(x\), \(y\), and \(z\) in terms of \(w\): - \(w = w\) - \(z = -w\) - \(y = 3w\) - \(x = 4w\) **Now substitute into equation (4) to check consistency:** \[ 2(4w) + 4(3w) + (-w) + 3w = 0 \] \[ 8w + 12w - w + 3w = 0 \] \[ (8 + 12 - 1 + 3)w = 0 \implies 22w = 0 \] This means \(w = 0\). Now, substituting \(w = 0\) into our expressions for \(x\), \(y\), and \(z\): - \(w = 0\) - \(z = -w = 0\) - \(y = 3w = 0\) - \(x = 4w = 0\) Finally, the solution to the system is: \[ (x, y, z, w) = (0, 0, 0, 0) \] Thus, the solution is \((0, 0, 0, 0)\).