1. **State the problem:** Solve the system of linear equations: $$3x + 8y = 34$$ $$5x + 6y = 20$$ 2. **Formula and method:** We will use the method of elimination to solve for $x$ and $y$. The goal is to eliminate one variable by making the coefficients of that variable equal in both equations. 3. **Multiply equations to align coefficients:** Multiply the first equation by 5 and the second by 3 to align the coefficients of $x$: $$5(3x + 8y) = 5(34) \Rightarrow 15x + 40y = 170$$ $$3(5x + 6y) = 3(20) \Rightarrow 15x + 18y = 60$$ 4. **Subtract the second equation from the first to eliminate $x$:** $$ (15x + 40y) - (15x + 18y) = 170 - 60 $$ $$ 15x - 15x + 40y - 18y = 110 $$ $$ 22y = 110 $$ 5. **Solve for $y$:** $$ y = \frac{110}{22} $$ Show cancellation: $$ y = \frac{\cancel{110}}{\cancel{22}} = 5 $$ 6. **Substitute $y=5$ into one of the original equations to find $x$:** Using the first equation: $$ 3x + 8(5) = 34 $$ $$ 3x + 40 = 34 $$ $$ 3x = 34 - 40 $$ $$ 3x = -6 $$ 7. **Solve for $x$:** $$ x = \frac{-6}{3} $$ Show cancellation: $$ x = \frac{\cancel{-6}}{\cancel{3}} = -2 $$ **Final answer:** $$ x = -2, \quad y = 5 $$