1. **State the problem:** Solve the equation $$\frac{2x + 1}{3y - 8} = \frac{x + 7}{y + 5}$$ for $x$ and $y$. 2. **Cross-multiply to eliminate the fractions:** $$ (2x + 1)(y + 5) = (x + 7)(3y - 8) $$ 3. **Expand both sides:** $$ 2x y + 10x + y + 5 = 3x y - 8x + 21 y - 56 $$ 4. **Group like terms:** Bring all terms to one side: $$ 2x y + 10x + y + 5 - 3x y + 8x - 21 y + 56 = 0 $$ Simplify: $$ (2x y - 3x y) + (10x + 8x) + (y - 21 y) + (5 + 56) = 0 $$ $$ -x y + 18x - 20 y + 61 = 0 $$ 5. **Rewrite the equation:** $$ -x y + 18x - 20 y + 61 = 0 $$ 6. **Solve for $x$ in terms of $y$:** $$ -x y + 18x = 20 y - 61 $$ $$ x(-y + 18) = 20 y - 61 $$ $$ x = \frac{20 y - 61}{-y + 18} = \frac{20 y - 61}{18 - y} $$ 7. **Alternatively, solve for $y$ in terms of $x$:** $$ -x y - 20 y = -18 x - 61 $$ $$ y(-x - 20) = -18 x - 61 $$ $$ y = \frac{-18 x - 61}{-x - 20} = \frac{18 x + 61}{x + 20} $$ **Final answer:** $$ x = \frac{20 y - 61}{18 - y} \quad \text{or} \quad y = \frac{18 x + 61}{x + 20} $$ This expresses $x$ in terms of $y$ or $y$ in terms of $x$ for the given equation.