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.)

\[ \begin{cases} x - 3y + 2z = -6 \\ 3x + y + 4z = 8 \\ 5x - 5y + 8z = -4 \end{cases} \]

\((x, y, z) = (\boxed{\quad})\)

Solution

Step 1: Write down the system of equations. \[ \begin{cases} x - 3y + 2z = -6 & \\ 3x + y + 4z = 8 & \\ 5x - 5y + 8z = -4 & \end{cases} \] Step 2: Eliminate \( x \) from the second and third equations using the first equation. Multiply the first equation by 3: \[ 3x - 9y + 6z = -18 & \] Subtract the second equation from this result: \[ (3x - 9y + 6z) - (3x + y + 4z) = -18 - 8 \\ -10y + 2z = -26 & \] Step 3: Eliminate \( x \) from the third equation using the first equation. Multiply the first equation by 5: \[ 5x - 15y + 10z = -30 & \] Subtract the third equation from this result: \[ (5x - 15y + 10z) - (5x - 5y + 8z) = -30 - (-4) \\ -10y + 2z = -26 & \] Step 4: Solve for \( y \) in terms of \( z \). \[ -10y + 2z = -26 \\ -10y = -26 - 2z \\ y = \frac{26 + 2z}{10} \\ y = \frac{13 + z}{5} & \] Step 5: Express \( z \) as a parameter \( c \). Let \( z = c \). Then, \[ y = \frac{13 + c}{5} & \] Step 6: Substitute \( y \) and \( z \) into the first equation to solve for \( x \). \[ x - 3\left(\frac{13 + c}{5}\right) + 2c = -6 \\ x - \frac{39 + 3c}{5} + 2c = -6 \\ x = -6 + \frac{39 + 3c}{5} - 2c \\ x = \frac{-30 + 39 + 3c - 10c}{5} \\ x = \frac{9 - 7c}{5} & \] The general solution is: \[ (x, y, z) = \left( \frac{9 - 7c}{5},\ \frac{13 + c}{5},\ c \right) \]