1. **State the problem:** Solve the system of linear equations using Gaussian elimination: $$\begin{cases} 2x_1 + 4x_2 + 5x_3 = 7 \\ x_1 + 3x_2 + x_3 = 10 \\ 3x_1 + 5x_2 + 3x_3 = 2 \end{cases}$$ 2. **Write the augmented matrix:** $$\left[\begin{array}{ccc|c} 2 & 4 & 5 & 7 \\ 1 & 3 & 1 & 10 \\ 3 & 5 & 3 & 2 \end{array}\right]$$ 3. **Goal:** Use row operations to convert the matrix to row echelon form (upper triangular), then back-substitute to find $x_1$, $x_2$, and $x_3$. 4. **Step 1:** Swap Row 1 and Row 2 to get a leading 1 in the first row: $$\left[\begin{array}{ccc|c} 1 & 3 & 1 & 10 \\ 2 & 4 & 5 & 7 \\ 3 & 5 & 3 & 2 \end{array}\right]$$ 5. **Step 2:** Eliminate $x_1$ terms in Rows 2 and 3: - Row 2 = Row 2 - 2 * Row 1: $$2 - 2(1) = 0, \quad 4 - 2(3) = 4 - 6 = -2, \quad 5 - 2(1) = 5 - 2 = 3, \quad 7 - 2(10) = 7 - 20 = -13$$ - Row 3 = Row 3 - 3 * Row 1: $$3 - 3(1) = 0, \quad 5 - 3(3) = 5 - 9 = -4, \quad 3 - 3(1) = 3 - 3 = 0, \quad 2 - 3(10) = 2 - 30 = -28$$ Matrix now: $$\left[\begin{array}{ccc|c} 1 & 3 & 1 & 10 \\ 0 & -2 & 3 & -13 \\ 0 & -4 & 0 & -28 \end{array}\right]$$ 6. **Step 3:** Make the pivot in Row 2 a 1 by dividing Row 2 by -2: $$\left[\begin{array}{ccc|c} 1 & 3 & 1 & 10 \\ 0 & 1 & -\frac{3}{2} & \frac{13}{2} \\ 0 & -4 & 0 & -28 \end{array}\right]$$ 7. **Step 4:** Eliminate the $x_2$ term in Row 3 using Row 2: Row 3 = Row 3 + 4 * Row 2: $$0, \quad -4 + 4(1) = 0, \quad 0 + 4(-\frac{3}{2}) = 0 - 6 = -6, \quad -28 + 4(\frac{13}{2}) = -28 + 26 = -2$$ Matrix now: $$\left[\begin{array}{ccc|c} 1 & 3 & 1 & 10 \\ 0 & 1 & -\frac{3}{2} & \frac{13}{2} \\ 0 & 0 & -6 & -2 \end{array}\right]$$ 8. **Step 5:** Make the pivot in Row 3 a 1 by dividing Row 3 by -6: $$\left[\begin{array}{ccc|c} 1 & 3 & 1 & 10 \\ 0 & 1 & -\frac{3}{2} & \frac{13}{2} \\ 0 & 0 & 1 & \frac{1}{3} \end{array}\right]$$ 9. **Step 6:** Back-substitute to eliminate $x_3$ terms in Rows 1 and 2: - Row 2 = Row 2 + $\frac{3}{2}$ * Row 3: $$0, 1, 0, \frac{13}{2} + \frac{3}{2} \times \frac{1}{3} = \frac{13}{2} + \frac{3}{6} = \frac{13}{2} + \frac{1}{2} = 7$$ - Row 1 = Row 1 - 1 * Row 3: $$1, 3, 0, 10 - 1 \times \frac{1}{3} = 10 - \frac{1}{3} = \frac{30}{3} - \frac{1}{3} = \frac{29}{3}$$ Matrix now: $$\left[\begin{array}{ccc|c} 1 & 3 & 0 & \frac{29}{3} \\ 0 & 1 & 0 & 7 \\ 0 & 0 & 1 & \frac{1}{3} \end{array}\right]$$ 10. **Step 7:** Eliminate $x_2$ term in Row 1: Row 1 = Row 1 - 3 * Row 2: $$1, 0, 0, \frac{29}{3} - 3(7) = \frac{29}{3} - 21 = \frac{29}{3} - \frac{63}{3} = -\frac{34}{3}$$ 11. **Final solution:** $$x_1 = -\frac{34}{3}, \quad x_2 = 7, \quad x_3 = \frac{1}{3}$$