1. **State the problem:** Solve the system of linear equations: $$-2x - 5y = -15$$ $$10x - 4y = 105$$ 2. **Formula and rules:** We can solve this system using the elimination or substitution method. Here, we'll use elimination to eliminate one variable. 3. **Multiply the first equation by 5** to align coefficients of $x$: $$5(-2x - 5y) = 5(-15)$$ $$-10x - 25y = -75$$ 4. **Add this to the second equation:** $$(-10x - 25y) + (10x - 4y) = -75 + 105$$ $$\cancel{-10x} - 25y + \cancel{10x} - 4y = 30$$ $$-29y = 30$$ 5. **Solve for $y$:** $$y = \frac{30}{-29} = -\frac{30}{29}$$ 6. **Substitute $y$ back into the first equation:** $$-2x - 5\left(-\frac{30}{29}\right) = -15$$ $$-2x + \frac{150}{29} = -15$$ 7. **Isolate $x$:** $$-2x = -15 - \frac{150}{29} = -\frac{435}{29} - \frac{150}{29} = -\frac{585}{29}$$ 8. **Divide both sides by $-2$:** $$x = \frac{-\frac{585}{29}}{-2} = \frac{585}{29 \times 2} = \frac{585}{58}$$ 9. **Simplify $x$ if possible:** $$585 = 9 \times 65, \quad 58 = 2 \times 29$$ No common factors, so $$x = \frac{585}{58}$$ **Final solution:** $$x = \frac{585}{58}, \quad y = -\frac{30}{29}$$