1. **Problem statement:** Solve the system of linear equations: $$3x + y = 9$$ $$6x + 7y = 5$$ 2. **Check if LU decomposition is applicable:** LU decomposition requires the coefficient matrix to be square and non-singular (determinant $\neq 0$). The coefficient matrix is: $$A = \begin{bmatrix} 3 & 1 \\ 6 & 7 \end{bmatrix}$$ Calculate determinant: $$\det(A) = 3 \times 7 - 6 \times 1 = 21 - 6 = 15 \neq 0$$ Since determinant is not zero, LU decomposition can be used. 3. **Solve the system by hand:** We use substitution or elimination. Here, elimination: Multiply the first equation by 7: $$7(3x + y) = 7 \times 9 \Rightarrow 21x + 7y = 63$$ Subtract the second equation: $$21x + 7y - (6x + 7y) = 63 - 5$$ Simplify: $$21x + 7y - 6x - 7y = 58$$ $$15x = 58$$ Solve for $x$: $$x = \frac{58}{15}$$ Substitute $x$ back into the first equation: $$3 \times \frac{58}{15} + y = 9$$ $$\frac{174}{15} + y = 9$$ $$y = 9 - \frac{174}{15} = \frac{135}{15} - \frac{174}{15} = -\frac{39}{15} = -\frac{13}{5}$$ **Final solution:** $$x = \frac{58}{15}, \quad y = -\frac{13}{5}$$