1. The problem is to solve a system of linear equations using the Gaussian elimination method. 2. Gaussian elimination involves transforming the system's augmented matrix into an upper triangular form using row operations. 3. The key row operations are swapping rows, multiplying a row by a nonzero scalar, and adding a multiple of one row to another. 4. After obtaining the upper triangular matrix, back substitution is used to find the solution. 5. For example, consider the system: $$\begin{cases} 2x + 3y = 8 \\ 4x + 9y = 20 \end{cases}$$ 6. Write the augmented matrix: $$\left[\begin{array}{cc|c} 2 & 3 & 8 \\ 4 & 9 & 20 \end{array}\right]$$ 7. Use row operations to eliminate the $x$ term in the second row: Multiply row 1 by $\frac{4}{2} = 2$: $$\left[\begin{array}{cc|c} 2 & 3 & 8 \\ \cancel{4} & 9 & 20 \end{array}\right] \to \left[\begin{array}{cc|c} 2 & 3 & 8 \\ 4 & 9 & 20 \end{array}\right]$$ Subtract row 1 multiplied by 2 from row 2: $$R_2 \to R_2 - 2R_1$$ $$\left[\begin{array}{cc|c} 2 & 3 & 8 \\ \cancel{4} & 9 & 20 \end{array}\right] - \left[\begin{array}{cc|c} 4 & 6 & 16 \end{array}\right] = \left[\begin{array}{cc|c} 2 & 3 & 8 \\ 0 & 3 & 4 \end{array}\right]$$ 8. Now the matrix is upper triangular: $$\left[\begin{array}{cc|c} 2 & 3 & 8 \\ 0 & 3 & 4 \end{array}\right]$$ 9. Back substitution: From second row: $3y = 4 \Rightarrow y = \frac{4}{3}$ From first row: $2x + 3\left(\frac{4}{3}\right) = 8 \Rightarrow 2x + 4 = 8 \Rightarrow 2x = 4 \Rightarrow x = 2$ 10. Final solution: $$\boxed{\left(x,y\right) = \left(2, \frac{4}{3}\right)}$$