1. We are given the system of equations: $$3x + 8y = 12 \quad (1)$$ $$5x + 7y = 1 \quad (2)$$ Our goal is to find the values of $x$ and $y$ that satisfy both equations simultaneously. 2. One common method to solve such systems is the **elimination method**. We aim to eliminate one variable by making the coefficients of $x$ or $y$ the same in both equations. 3. Multiply equation (1) by 5 and equation (2) by 3 to align the coefficients of $x$: $$5 \times (3x + 8y) = 5 \times 12 \Rightarrow 15x + 40y = 60$$ $$3 \times (5x + 7y) = 3 \times 1 \Rightarrow 15x + 21y = 3$$ 4. Subtract the second new equation from the first to eliminate $x$: $$ (15x + 40y) - (15x + 21y) = 60 - 3 $$ $$ 15x - 15x + 40y - 21y = 57 $$ $$ 19y = 57 $$ 5. Solve for $y$: $$ y = \frac{57}{19} = 3 $$ 6. Substitute $y=3$ back into one of the original equations, say equation (1): $$ 3x + 8(3) = 12 $$ $$ 3x + 24 = 12 $$ $$ 3x = 12 - 24 = -12 $$ $$ x = \frac{-12}{3} = -4 $$ 7. The solution to the system is the ordered pair: $$ \boxed{(-4, 3)} $$ 8. Since we found a unique solution, the system is consistent and independent.