1. **State the problem:** We are given the system of linear equations: $$\begin{cases} x_1 + x_2 - 5x_3 = -8 \\ 5x_1 + 4x_2 - 5x_3 = -8 \end{cases}$$ and we want to express the solution vector $\mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}$ in the form: $$\mathbf{x} = \mathbf{p} + s \mathbf{v}$$ where $\mathbf{p}$ and $\mathbf{v}$ are 3x1 vectors and $s$ is a scalar parameter. 2. **Rewrite the system:** Let $x_3 = s$ be the free parameter since there are two equations and three unknowns. 3. **Substitute $x_3 = s$ into the equations:** $$\begin{cases} x_1 + x_2 - 5s = -8 \\ 5x_1 + 4x_2 - 5s = -8 \end{cases}$$ 4. **Rewrite the system in terms of $x_1$ and $x_2$:** $$\begin{cases} x_1 + x_2 = -8 + 5s \\ 5x_1 + 4x_2 = -8 + 5s \end{cases}$$ 5. **Solve the system for $x_1$ and $x_2$:** Multiply the first equation by 4: $$4x_1 + 4x_2 = 4(-8 + 5s) = -32 + 20s$$ Subtract the second equation from this: $$\cancel{4x_1} + 4x_2 - (5x_1 + 4x_2) = (-32 + 20s) - (-8 + 5s)$$ Simplify left side: $$4x_1 + 4x_2 - 5x_1 - 4x_2 = -x_1$$ Simplify right side: $$-32 + 20s + 8 - 5s = -24 + 15s$$ So: $$-x_1 = -24 + 15s \implies x_1 = 24 - 15s$$ 6. **Find $x_2$ using the first equation:** $$x_1 + x_2 = -8 + 5s$$ Substitute $x_1$: $$24 - 15s + x_2 = -8 + 5s$$ Solve for $x_2$: $$x_2 = -8 + 5s - 24 + 15s = -32 + 20s$$ 7. **Write the solution vector:** $$\mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 24 - 15s \\ -32 + 20s \\ s \end{bmatrix} = \begin{bmatrix} 24 \\ -32 \\ 0 \end{bmatrix} + s \begin{bmatrix} -15 \\ 20 \\ 1 \end{bmatrix}$$ **Final answer:** $$\mathbf{x} = \begin{bmatrix} 24 \\ -32 \\ 0 \end{bmatrix} + s \begin{bmatrix} -15 \\ 20 \\ 1 \end{bmatrix}$$ This expresses the solution as a particular solution plus a scalar multiple of a vector spanning the solution space.