1. **State the problem:** Solve the system of linear equations: $$2x + 3y = 13$$ $$x - 4y = -5$$ 2. **Formula and method:** We can use substitution or elimination. Here, we'll use substitution. 3. **Isolate $x$ in the second equation:** $$x = -5 + 4y$$ 4. **Substitute $x$ into the first equation:** $$2(-5 + 4y) + 3y = 13$$ 5. **Simplify and solve for $y$:** $$-10 + 8y + 3y = 13$$ $$11y - 10 = 13$$ $$11y = 23$$ $$y = \frac{23}{11}$$ 6. **Substitute $y$ back to find $x$:** $$x = -5 + 4 \times \frac{23}{11} = -5 + \frac{92}{11} = \frac{-55}{11} + \frac{92}{11} = \frac{37}{11}$$ 7. **Final answer:** $$x = \frac{37}{11}, \quad y = \frac{23}{11}$$