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

Solve the system.

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

\(x = \)

\(y = \)

\(z = \)

Solution

To solve the system of equations, we will use the method of elimination or substitution. The given system is: \[ \begin{cases} 2x + 2y - z = 7 & \\ -x - 2y + 2z = -6 & \\ -2x - 4y + 3z = -9 & \end{cases} \] First, we’ll eliminate one of the variables. Let’s eliminate \(y\) from the first two equations. Multiply the first equation by 1: \[ 2x + 2y - z = 7 \] Multiply the second equation by 1: \[ -x - 2y + 2z = -6 \] Add these two equations to eliminate \(y\): \[ (2x + 2y - z) + (-x - 2y + 2z) = 7 - 6 \] Simplify: \[ x + z = 1 \] Now we have: \[ x + z = 1 \] Next, eliminate \(y\) from the second and third equations. Multiply the second equation by 2: \[ -2x - 4y + 4z = -12 \] Leave the third equation as it is: \[ -2x - 4y + 3z = -9 \] Subtract the third equation from the modified second equation to eliminate \(y\): \[ (-2x - 4y + 4z) - (-2x - 4y + 3z) = -12 + 9 \] Simplify: \[ z = -3 \] Now, we substitute \(z = -3\) back into the equation \(x + z = 1\): \[ x - 3 = 1 \] So, \[ x = 4 \] Now substitute \(x = 4\) and \(z = -3\) into the first original equation: \[ 2(4) + 2y - (-3) = 7 \] Simplify: \[ 8 + 2y + 3 = 7 \] \[ 2y + 11 = 7 \] \[ 2y = -4 \] \[ y = -2 \] The solution to the system of equations is: \[ x = 4, \quad y = -2, \quad z = -3 \]