1. **State the problem:** Solve the system of equations: $$x - 4y = -5$$ $$-3x + 5y = 1$$ 2. **Write the system in tabular (augmented matrix) form:** $$\begin{bmatrix} 1 & -4 & | & -5 \\ -3 & 5 & | & 1 \end{bmatrix}$$ 3. **Use row operations to solve:** - Start with the matrix: $$\begin{bmatrix} 1 & -4 & | & -5 \\ -3 & 5 & | & 1 \end{bmatrix}$$ - Add 3 times row 1 to row 2 to eliminate $x$ from row 2: $$R_2 \to R_2 + 3R_1$$ $$\begin{bmatrix} 1 & -4 & | & -5 \\ \cancel{-3} + 3 \times 1 & 5 + 3 \times (-4) & | & 1 + 3 \times (-5) \end{bmatrix} = \begin{bmatrix} 1 & -4 & | & -5 \\ 0 & -7 & | & -14 \end{bmatrix}$$ - Divide row 2 by $-7$ to make the pivot 1: $$R_2 \to \frac{1}{-7} R_2$$ $$\begin{bmatrix} 1 & -4 & | & -5 \\ 0 & \cancel{-7} \times \frac{1}{-7} & | & \cancel{-14} \times \frac{1}{-7} \end{bmatrix} = \begin{bmatrix} 1 & -4 & | & -5 \\ 0 & 1 & | & 2 \end{bmatrix}$$ - Add 4 times row 2 to row 1 to eliminate $y$ from row 1: $$R_1 \to R_1 + 4R_2$$ $$\begin{bmatrix} 1 & \cancel{-4} + 4 \times 1 & | & -5 + 4 \times 2 \\ 0 & 1 & | & 2 \end{bmatrix} = \begin{bmatrix} 1 & 0 & | & 3 \\ 0 & 1 & | & 2 \end{bmatrix}$$ 4. **Solution:** $$x = 3, \quad y = 2$$ Thus, the exact solution is $(3, 2)$.