Question: Solve the system. \[ \begin{align*} 3x + y - z &= 3 \\ -x + 2y + z &= 0 \\ -5x + 2y + 3z &=…

Solve the system.

\[ \begin{align*} 3x + y - z &= 3 \\ -x + 2y + z &= 0 \\ -5x + 2y + 3z &= -2 \end{align*} \]

\( x = \boxed{\phantom{0}} \)

\( y = \boxed{\phantom{0}} \)

\( z = \boxed{\phantom{0}} \)

Solution

To solve the system of equations, we’ll use the elimination or substitution method. The system is: \[ \begin{cases} 3x + y - z = 3 & \\ -x + 2y + z = 0 & \\ -5x + 2y + 3z = -2 & \end{cases} \] Step 1: We can add the first two equations to eliminate \( z \): \[ (3x + y - z) + (-x + 2y + z) = 3 + 0 \] \[ 2x + 3y = 3 \] Step 2: We’ll eliminate \( z \) between the second and third equations: \[ (-x + 2y + z) + (-5x + 2y + 3z) = 0 - 2 \] \[ -6x + 4y + 4z = -2 \] Divide the entire equation by 2: \[ -3x + 2y + 2z = -1 \] Step 3: Now substitute \( z = (2x + 3y - 3)/1 \) from the simplified equation \( 2x + 3y = 3 \). Step 4: Substitute back into one of the original equations to solve for \( y \). Using the first equation: \[ 3x + y - \left(\frac{2x + 3y - 3}{1}\right) = 3 \] Simplify: \[ 3x + y - 2x - 3y + 3 = 3 \] Combine like terms: \[ x - 2y + 3 = 3 \] \[ x - 2y = 0 \] From Step 2 and Step 3 solve for \( y \): \[ x = 2y \] Step 5: Substitute \( x = 2y \) into one of the equations: Substituting in \( 2x + 3y = 3 \): \[ 2(2y) + 3y = 3 \] \[ 4y + 3y = 3 \] \[ 7y = 3 \] \[ y = \frac{3}{7} \] Step 6: \( x = 2y = 2\left(\frac{3}{7}\right) = \frac{6}{7} \). Step 7: Calculate \( z \) using \( x \) and \( y \) from Equation 1: \[ 3\left(\frac{6}{7}\right) + \frac{3}{7} - z = 3 \] Simplify and solve for \( z \): \[ \frac{18}{7} + \frac{3}{7} - z = 3 \] \[ \frac{21}{7} - z = 3 \] \[ z = \frac{21}{7} - 3 = 0 \] The solution is: \[ x = \frac{6}{7}, \quad y = \frac{3}{7}, \quad z = 0 \]