To solve this system of equations, we can use either the elimination method or matrix methods (such as Gaussian elimination). Let's go step-by-step with Gaussian elimination.

Given: $2x - 3y - 6z = 35$ $-9x + 5y - z = 8$ $-5x + 8y - 7z = 21$

Step 1: Write the system in augmented matrix form.

2 & -3 & -6 & | & 35 \\ -9 & 5 & -1 & | & 8 \\ -5 & 8 & -7 & | & 21 \\ \end{bmatrix}$$ ### Step 2: Use row operations to get a row-echelon form. 1. **Make the leading coefficient of the first row equal to 1** by dividing the first row by 2: $$R_1: \quad \left[1, -\frac{3}{2}, -3 \; | \; 17.5 \right]$$ 2. **Eliminate the $$x$$-term in rows 2 and 3**: - For $$R_2$$: $$R_2 + 9 \times R_1 \rightarrow R_2$$ - For $$R_3$$: $$R_3 + 5 \times R_1 \rightarrow R_3$$ After performing these operations, we get: $$\begin{bmatrix} 1 & -\frac{3}{2} & -3 & | & 17.5 \\ 0 & \frac{1}{2} & 26 & | & 166 \\ 0 & \frac{5}{2} & -22 & | & 108 \\ \end{bmatrix}$$ 3. **Make the leading coefficient of the second row equal to 1** by multiplying $$R_2$$ by 2: $$R_2: \quad \left[0, 1, 52 \; | \; 332 \right]$$ 4. **Eliminate the $$y$$-term in row 3** by performing $$R_3 - \frac{5}{2} \times R_2 \rightarrow R_3$$: $$\begin{bmatrix} 1 & -\frac{3}{2} & -3 & | & 17.5 \\ 0 & 1 & 52 & | & 332 \\ 0 & 0 & -282 & | & -722 \\ \end{bmatrix}$$ 5. **Solve for $$z$$** using the third row: $$-282z = -722 \Rightarrow z = \frac{722}{282} = \frac{361}{141} = 2.5$$ 6. **Back-substitute $$z = 2.5$$ into $$R_2$$ and $$R_1$$ to find $$y$$ and $$x$$**: - From $$R_2$$: $$y + 52 \times 2.5 = 332$$ - Solve for $$y$$: $$y = 332 - 130 = 202$$ - Substitute $$y = 202$$ and $$z = 2.5$$ into $$R_1$$: $$x - \frac{3}{2} \times 202 - 3 \times 2.5 = 17.5$$ Solving this, we get $$x$$. After calculating