1. **State the problem:** Solve the system of equations: $$7x - 8 = -5y$$ $$3x + y = 4$$ 2. **Rewrite the first equation to express $y$ in terms of $x$: ** $$7x - 8 = -5y \implies -5y = 7x - 8 \implies y = \frac{8 - 7x}{5}$$ 3. **Substitute $y$ from the first equation into the second equation:** $$3x + y = 4 \implies 3x + \frac{8 - 7x}{5} = 4$$ 4. **Multiply both sides by 5 to clear the denominator:** $$5 \times \left(3x + \frac{8 - 7x}{5}\right) = 5 \times 4$$ $$5 \times 3x + \cancel{5} \times \frac{8 - 7x}{\cancel{5}} = 20$$ $$15x + 8 - 7x = 20$$ 5. **Simplify the left side:** $$15x - 7x + 8 = 20 \implies 8x + 8 = 20$$ 6. **Subtract 8 from both sides:** $$8x + 8 - 8 = 20 - 8 \implies 8x = 12$$ 7. **Divide both sides by 8:** $$\frac{\cancel{8}x}{\cancel{8}} = \frac{12}{8} \implies x = \frac{12}{8} = \frac{3}{2}$$ 8. **Substitute $x = \frac{3}{2}$ back into the expression for $y$:** $$y = \frac{8 - 7 \times \frac{3}{2}}{5} = \frac{8 - \frac{21}{2}}{5} = \frac{\frac{16}{2} - \frac{21}{2}}{5} = \frac{-\frac{5}{2}}{5} = -\frac{5}{2} \times \frac{1}{5} = -\frac{1}{2}$$ 9. **Final solution:** $$\boxed{\left(\frac{3}{2}, -\frac{1}{2}\right)}$$ This is the point where the two lines intersect on the coordinate plane.