1. The problem is to solve the system of linear equations: $$6x - 5y = 9$$ $$8x + 11y = 20$$ $$5x - 11y = -59$$ 2. We will use the method of elimination or substitution to find values of $x$ and $y$ that satisfy all three equations. 3. First, observe the second and third equations: $$8x + 11y = 20$$ $$5x - 11y = -59$$ Add these two equations to eliminate $y$: $$ (8x + 11y) + (5x - 11y) = 20 + (-59) $$ $$ 8x + 5x + \cancel{11y} - \cancel{11y} = -39 $$ $$ 13x = -39 $$ 4. Solve for $x$: $$ x = \frac{-39}{13} = -3 $$ 5. Substitute $x = -3$ into the second equation to find $y$: $$ 8(-3) + 11y = 20 $$ $$ -24 + 11y = 20 $$ $$ 11y = 20 + 24 = 44 $$ $$ y = \frac{44}{11} = 4 $$ 6. Verify the solution $(x,y) = (-3,4)$ in the first equation: $$ 6(-3) - 5(4) = -18 - 20 = -38 $$ The left side equals $-38$, but the right side is $9$, so the solution does not satisfy the first equation. 7. Since the three equations are inconsistent, there is no solution that satisfies all three simultaneously. Final answer: The system of equations has no solution (it is inconsistent).