Question: Solve the following system. \[ \left\{ \begin{array}{l} 2x - y + 3z = 5 \\ 3x + 4y - 3z = -6…

Solve the following system.

\[ \left\{ \begin{array}{l} 2x - y + 3z = 5 \\ 3x + 4y - 3z = -6 \\ x + 2z = 6 \end{array} \right. \]

\[ x = \boxed{\phantom{0}} \]

\[ y = \boxed{\phantom{0}} \]

\[ z = \boxed{\phantom{0}} \]

Solution

To solve the system of equations: \[ \begin{cases} 2x - y + 3z = 5 & \\ 3x + 4y - 3z = -6 & \\ x + 2z = 6 & \end{cases} \] First, solve the third equation for \(x\): \[ x + 2z = 6 \] Subtract \(2z\) from both sides: \[ x = 6 - 2z \] Substitute \(x = 6 - 2z\) into the first equation: \[ 2(6 - 2z) - y + 3z = 5 \] Simplify: \[ 12 - 4z - y + 3z = 5 \] Combine like terms: \[ 12 - z - y = 5 \] Subtract 12 from both sides: \[ -z - y = -7 \] Multiply the entire equation by \(-1\): \[ z + y = 7 \] Now substitute \(x = 6 - 2z\) into the second equation: \[ 3(6 - 2z) + 4y - 3z = -6 \] Simplify: \[ 18 - 6z + 4y - 3z = -6 \] Combine like terms: \[ 18 - 9z + 4y = -6 \] Subtract 18 from both sides: \[ -9z + 4y = -24 \] Simplify the equation: \[ 3z - \frac{4}{3}y = 8 \] Now solve the system for \(z\) and \(y\): Substitute \(y = 7 - z\) from the equation \(z + y = 7\) into the second simplified equation: \[ 3z - \frac{4}{3}(7 - z) = 8 \] Simplify: \[ 3z - \frac{28}{3} + \frac{4}{3}z = 8 \] Combine like terms: \[ \frac{9z}{3} + \frac{4z}{3} = 8 + \frac{28}{3} \] \[ \frac{13z}{3} = \frac{24}{3} \] Solve for \(z\): \[ 13z = 24 \] \[ z = \frac{24}{13} \] Substitute \(z = \frac{24}{13}\) into \(y = 7 - z\): \[ y = 7 - \frac{24}{13} \] Convert 7 to a fraction with denominator 13: \[ y = \frac{91}{13} - \frac{24}{13} \] \[ y = \frac{67}{13} \] Finally, substitute \(z = \frac{24}{13}\) back into \(x = 6 - 2z\): \[ x = 6 - 2 \times \frac{24}{13} \] Convert 6 to a fraction with denominator 13: \[ x = \frac{78}{13} - \frac{48}{13} \] \[ x = \frac{30}{13} \] The solution to the system is: \[ x = \frac{30}{13}, \quad y = \frac{67}{13}, \quad z = \frac{24}{13} \]