1. **State the problem:** Solve the system of linear equations: $$2x + 3y = 7$$ $$2x = 4y - 5$$ 2. **Rewrite the second equation:** $$2x = 4y - 5 \implies 2x - 4y = -5$$ 3. **Subtract the second equation from the first:** $$\left(2x + 3y\right) - \left(2x - 4y\right) = 7 - (-5)$$ Simplify: $$2x + 3y - 2x + 4y = 7 + 5$$ $$7y = 12$$ 4. **Solve for $y$:** $$y = \frac{12}{7}$$ 5. **Substitute $y$ back into the first equation:** $$2x + 3\left(\frac{12}{7}\right) = 7$$ $$2x + \frac{36}{7} = 7$$ 6. **Isolate $x$:** $$2x = 7 - \frac{36}{7} = \frac{49}{7} - \frac{36}{7} = \frac{13}{7}$$ 7. **Solve for $x$:** $$x = \frac{\cancel{2} \times x}{\cancel{2}} = \frac{13}{7} \times \frac{1}{2} = \frac{13}{14}$$ **Final answer:** $$x = \frac{13}{14}, \quad y = \frac{12}{7}$$