1. **Problem Statement:** Let's understand what "Systems of Linear Equations" and "Gaussian Elimination" mean in simple language. 2. **Systems of Linear Equations:** A system of linear equations is a set of two or more equations where each equation is linear. For example: $$\begin{cases} 2x + 3y = 6 \\ x - y = 1 \end{cases}$$ Here, $x$ and $y$ are variables, and the goal is to find values of $x$ and $y$ that satisfy both equations at the same time. 3. **Why solve systems?** Because many real-world problems involve multiple conditions that must be true simultaneously, and these can be modeled as systems of linear equations. 4. **Gaussian Elimination:** This is a step-by-step method to solve systems of linear equations by transforming the system into a simpler form called "row echelon form" using operations like swapping equations, multiplying an equation by a number, or adding/subtracting equations. 5. **Steps of Gaussian Elimination:** - Write the system as an augmented matrix. - Use row operations to create zeros below the main diagonal. - Once in row echelon form, use back substitution to find the values of variables. 6. **Example:** Solve the system: $$\begin{cases} x + 2y = 5 \\ 3x + 4y = 11 \end{cases}$$ 7. **Write augmented matrix:** $$\left[\begin{array}{cc|c} 1 & 2 & 5 \\ 3 & 4 & 11 \end{array}\right]$$ 8. **Make zero below first pivot (1st row, 1st column):** Replace row 2 by row 2 minus 3 times row 1: $$\left[\begin{array}{cc|c} 1 & 2 & 5 \\ 3 - 3\times1 & 4 - 3\times2 & 11 - 3\times5 \end{array}\right] = \left[\begin{array}{cc|c} 1 & 2 & 5 \\ 0 & \cancel{4} - 6 & \cancel{11} - 15 \end{array}\right] = \left[\begin{array}{cc|c} 1 & 2 & 5 \\ 0 & -2 & -4 \end{array}\right]$$ 9. **Make pivot 1 in second row:** Multiply row 2 by $\frac{1}{-2}$: $$\left[\begin{array}{cc|c} 1 & 2 & 5 \\ 0 & \cancel{-2} \times \frac{1}{\cancel{-2}} & -4 \times \frac{1}{-2} \end{array}\right] = \left[\begin{array}{cc|c} 1 & 2 & 5 \\ 0 & 1 & 2 \end{array}\right]$$ 10. **Eliminate above second pivot:** Replace row 1 by row 1 minus 2 times row 2: $$\left[\begin{array}{cc|c} 1 & 2 - 2\times1 & 5 - 2\times2 \\ 0 & 1 & 2 \end{array}\right] = \left[\begin{array}{cc|c} 1 & \cancel{2} - 2 & 5 - 4 \\ 0 & 1 & 2 \end{array}\right] = \left[\begin{array}{cc|c} 1 & 0 & 1 \\ 0 & 1 & 2 \end{array}\right]$$ 11. **Solution:** From the matrix, $x = 1$ and $y = 2$. **Final answer:** $$x = 1, \quad y = 2$$ This is how Gaussian Elimination helps solve systems of linear equations step by step.