1. **State the problem:** Solve the system of equations using the elimination method: $$3x + 2y = 18$$ $$4x = 7$$ 2. **Rewrite the second equation:** $$4x = 7 \implies x = \frac{7}{4}$$ 3. **Substitute $x = \frac{7}{4}$ into the first equation:** $$3\left(\frac{7}{4}\right) + 2y = 18$$ 4. **Simplify the substitution:** $$\frac{21}{4} + 2y = 18$$ 5. **Isolate $2y$ by subtracting $\frac{21}{4}$ from both sides:** $$2y = 18 - \frac{21}{4}$$ 6. **Convert 18 to quarters to subtract:** $$18 = \frac{72}{4}$$ So, $$2y = \frac{72}{4} - \frac{21}{4} = \frac{51}{4}$$ 7. **Divide both sides by 2 to solve for $y$:** $$y = \frac{\cancel{2} \times y}{\cancel{2}} = \frac{51}{4 \times 2} = \frac{51}{8}$$ 8. **Final solution:** $$x = \frac{7}{4}, \quad y = \frac{51}{8}$$ This means the solution to the system is $\left(\frac{7}{4}, \frac{51}{8}\right)$.