d. \( \begin{aligned} x+y+2 z & =4 \\ 2 x-4 y-z & =-14 \\ 3 x-2 y+z & =3\end{aligned} \)

To solve the system of equations given by: \[ \begin{aligned} 1) & \quad x + y + 2z = 4 \\ 2) & \quad 2x - 4y - z = -14 \\ 3) & \quad 3x - 2y + z = 3 \end{aligned} \] let's use the elimination or substitution method. Here's a step-by-step approach: First, let's isolate \( z \) in the first equation: \[ z = \frac{4 - x - y}{2} \] Substituting \( z \) into the second and third equations: Substituting into the second equation: \[ 2x - 4y - \left(\frac{4 - x - y}{2}\right) = -14 \] Multiplying through by 2 to eliminate the fraction: \[ 4x - 8y - (4 - x - y) = -28 \] This simplifies to: \[ 4x - 8y - 4 + x + y = -28 \\ 5x - 7y - 4 = -28 \\ 5x - 7y = -24 \] Now, substitute \( z \) into the third equation: \[ 3x - 2y + \left(\frac{4 - x - y}{2}\right) = 3 \] Again multiplying through by 2: \[ 6x - 4y + (4 - x - y) = 6 \\ 6x - 4y + 4 - x - y = 6 \\ 5x - 5y + 4 = 6 \\ 5x - 5y = 2 \\ x - y = \frac{2}{5} \Rightarrow x = y + \frac{2}{5} \] Now we have: 1) \( 5x - 7y = -24 \) 2) \( x = y + \frac{2}{5} \) Substituting \( x = y + \frac{2}{5} \) back into \( 5x - 7y = -24 \): \[ 5\left(y + \frac{2}{5}\right) - 7y = -24 \\ 5y + 2 - 7y = -24 \\ -2y + 2 = -24 \\ -2y = -26 \\ y = 13 \] Using \( y = 13 \) to find \( x \): \[ x = 13 + \frac{2}{5} = \frac{65}{5} + \frac{2}{5} = \frac{67}{5} \] Now to find \( z \) using \( y \) in the first equation: \[ x + 13 + 2z = 4 \\ \frac{67}{5} + 13 + 2z = 4 \] Convert 13 to a fraction: \[ \frac{67}{5} + \frac{65}{5} + 2z = 4 \\ \frac{132}{5} + 2z = 4 \\ 2z = 4 - \frac{132}{5} = \frac{20}{5} - \frac{132}{5} = -\frac{112}{5} \\ z = -\frac{56}{5} \] So, the solution to the system is: \[ \left( x, y, z \right) = \left( \frac{67}{5}, 13, -\frac{56}{5} \right) \]