Question: Solve the system of equations. (If the system is dependent, enter a general solution in terms of…

Solve the system of equations. (If the system is dependent, enter a general solution in terms of \( c \). If there is no solution, enter NO SOLUTION.)

\[ \begin{cases} 2x - 3y + 6z = 3 \\ x + 2y - 4z = 2 \\ 3x + 4y - 8z = 9 \end{cases} \]

\((x, y, z) = (\boxed{\phantom{0}}, \boxed{\phantom{0}}, \boxed{\phantom{0}})\)

Solution

Start by expressing \( x \) from the second equation: \[ x + 2y - 4z = 2 & \] \[ x = 2 - 2y + 4z & \] Substitute \( x \) into the first and third equations: \[ 2x - 3y + 6z = 3 & \] \[ 2(2 - 2y + 4z) - 3y + 6z = 3 & \] \[ 4 - 4y + 8z - 3y + 6z = 3 & \] \[ -7y + 14z = -1 & \] Divide by -7: \[ y - 2z = \frac{1}{7} & \] Now substitute \( x \) into the third equation: \[ 3x + 4y - 8z = 9 & \] \[ 3(2 - 2y + 4z) + 4y - 8z = 9 & \] \[ 6 - 6y + 12z + 4y - 8z = 9 & \] \[ -2y + 4z = 3 & \] Divide by -2: \[ y - 2z = -\frac{3}{2} & \] Now we have: \[ y - 2z = \frac{1}{7} & \] \[ y - 2z = -\frac{3}{2} & \] These equations are contradictory. Therefore, the system has no solution. NO SOLUTION