Answer
$\{(-7,8)\}$
Work Step by Step
1. Let $A=\begin{bmatrix} 2 & 1 \\ 3 & -1 \end{bmatrix}$ , we have $\begin{bmatrix} 2 & 1 \\ 3 & -1 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}=\begin{bmatrix} -6 \\ -29 \end{bmatrix}$
2. Thus $\begin{bmatrix} x \\ y \end{bmatrix}=A^{-1}\begin{bmatrix} -6 \\ -29 \end{bmatrix}$
3. Find $A^{-1}$ using row operations: $3R1-2R2\to R2, 5R1-R2\to R1, R1/10\to R1, R2/5\to R2$
$AI=\begin{bmatrix} 2 & 1 \\ 3 & -1 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} 2 & 1 \\ 0 & 5 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 3 & -2 \end{bmatrix}=\begin{bmatrix} 10 & 0 \\ 0 & 5 \end{bmatrix}\begin{bmatrix} 2& 2 \\ 3& -2 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 1/5 & 1/5 \\ 3/5 & -2/5 \end{bmatrix}$
4. Thus $A^{-1}=\begin{bmatrix} 1/5 & 1/5 \\ 3/5 & -2/5 \end{bmatrix}$
5. $\begin{bmatrix} x \\ y \end{bmatrix}=\begin{bmatrix} 1/5 & 1/5 \\ 3/5 & -2/5 \end{bmatrix}\begin{bmatrix} -6 \\ -29 \end{bmatrix}=\begin{bmatrix} -7 \\ 8 \end{bmatrix}$
6. Thus the solution set is $\{(-7,8)\}$
