1. **State the problem:** Solve the system of equations using substitution: $$x + 2y = 3$$ $$5x + 4y = 8$$ 2. **Isolate one variable:** From the first equation, solve for $x$: $$x = 3 - 2y$$ 3. **Substitute into the second equation:** Replace $x$ in the second equation with $3 - 2y$: $$5(3 - 2y) + 4y = 8$$ 4. **Simplify and solve for $y$:** $$15 - 10y + 4y = 8$$ $$15 - 6y = 8$$ 5. **Isolate $y$:** $$-6y = 8 - 15$$ $$-6y = -7$$ $$y = \frac{\cancel{-7}}{\cancel{-6}} = \frac{7}{6}$$ 6. **Substitute $y$ back to find $x$:** $$x = 3 - 2\left(\frac{7}{6}\right) = 3 - \frac{14}{6} = 3 - \frac{7}{3}$$ $$x = \frac{9}{3} - \frac{7}{3} = \frac{2}{3}$$ 7. **Final solution:** $$x = \frac{2}{3}, \quad y = \frac{7}{6}$$ This means the two lines intersect at the point $\left(\frac{2}{3}, \frac{7}{6}\right)$.