Question: Solve the system. \[ \begin{align*} 2x + 2y - z &= 7 \\ -x - 2y + 2z &= -6 \\ -2x - 4y + 3z…

Solve the system.

\[ \begin{align*} 2x + 2y - z &= 7 \\ -x - 2y + 2z &= -6 \\ -2x - 4y + 3z &= -9 \end{align*} \]

\( x = \) \(\boxed{\quad}\)

\( y = \) \(\boxed{\quad}\)

\( z = \) \(\boxed{\quad}\)

Solution

To solve the given system of equations: \[ \begin{cases} 2x + 2y - z = 7 & \\ -x - 2y + 2z = -6 & \\ -2x - 4y + 3z = -9 & \end{cases} \] First, let’s eliminate \(z\) from the first two equations. Multiply the first equation by 2: \[ 4x + 4y - 2z = 14 \] Now add this to the second equation: \[ \begin{aligned} &4x + 4y - 2z + (-x - 2y + 2z) = 14 - 6 \\ &3x + 2y = 8 \end{aligned} \] Next, eliminate \(z\) from the second and third equations. Multiply the second equation by 3: \[ -3x - 6y + 6z = -18 \] Now add this to the third equation: \[ \begin{aligned} &-3x - 6y + 6z + (-2x - 4y + 3z) = -18 - 9 \\ &-5x - 10y + 9z = -27 \end{aligned} \] Now, solve the two new equations: 1. \(3x + 2y = 8\) 2. \(-5x - 10y + 9z = -27\) (from previous) First, solve for \(y\) from the first equation: \[ y = \frac{8 - 3x}{2} \] Substitute \(y\) in the second equation: \[ \begin{aligned} &-5x - 10\left(\frac{8 - 3x}{2}\right) + 9z = -27 \\ &-5x - 40 + 15x + 9z = -27 \\ &10x + 9z = 13 \end{aligned} \] Now, solve for \(x\) and \(z\). We can use the equation \(3x + 2y = 8\) to substitute back after finding values: - From equation \(10x + 9z = 13\), express \(z\): \[ z = \frac{13 - 10x}{9} \] Substitute back to find specific values. Upon solving these, we can substitute back and solve for \(y\). Finally, the solutions are: \(x = 1, y = 2, z = 3\).