Solve the system of linear equations, if possible. State any solutions and classify each system as consistent independent, consistent dependent, or inconsistent. If not possible, enter \( D N E \) in part (b). If the system is dependent, enter \( x \) and \( y \) in terms of \( z \). \( x+y+z=-4 \) \( -3 x-y+z=0 \) \( 5 x+4 y+3 z=-14 \) (a) The system of equations is: Consistent dependent. Inconsistent. Consistent independent. (b) The solution to the system is \( x= \) or \( x= \)

**Step 1. Express one variable in terms of the others.** From the first equation, \[ x+y+z=-4, \] solve for \( z \): \[ z=-4-x-y. \] **Step 2. Substitute into the second equation.** The second equation is \[ -3x-y+z=0. \] Substitute \( z=-4-x-y \): \[ -3x-y+(-4-x-y)=0. \] Simplify: \[ -3x-y-4-x-y=0, \] \[ -4x-2y-4=0. \] Add 4 to both sides: \[ -4x-2y=4. \] Multiply by \(-1\): \[ 4x+2y=-4. \] Divide by 2: \[ 2x+y=-2. \] Solve for \( y \): \[ y=-2-2x. \] **Step 3. Express \( z \) using the value of \( y \).** Recall \[ z=-4-x-y. \] Substitute \( y=-2-2x \): \[ z=-4-x-(-2-2x)= -4-x+2+2x. \] Combine like terms: \[ z=(-4+2)+( -x+2x)= -2+x. \] Thus, \[ z=x-2. \] **Step 4. Substitute \( y \) and \( z \) into the third equation to verify consistency.** The third equation is \[ 5x+4y+3z=-14. \] Substitute \( y=-2-2x \) and \( z=x-2 \): \[ 5x+4(-2-2x)+3(x-2)=-14. \] Simplify: \[ 5x-8-8x+3x-6=-14. \] Combine like terms: \[ (5x-8x+3x) + (-8-6) = 0x -14=-14. \] This simplifies to: \[ -14=-14, \] which is always true. Hence, the third equation does not add any new information. **Step 5. Classify the system.** Since the third equation does not provide new information and the other two equations are consistent, the system has infinitely many solutions. It is a **consistent dependent** system. **Step 6. Express the solution in terms of \( z \).** We have: \[ z=x-2 \quad \Longrightarrow \quad x=z+2. \] And from earlier: \[ y=-2-2x. \] Substitute \( x=z+2 \): \[ y=-2-2(z+2)=-2-2z-4=-2z-6. \] **Final Answers:** (a) The system of equations is: **Consistent dependent.** (b) The solution to the system is \[ x=z+2,\quad y=-2z-6,\quad z=z\quad (\text{where } z \in \mathbb{R}). \]