1. **State the problem:** Solve the system of linear equations: $$3x - y = -4$$ $$\frac{1}{4}x - \frac{1}{3}y = \frac{7}{4}$$ 2. **Use substitution or elimination method.** Here, we use elimination. 3. Multiply the second equation by 12 (the least common multiple of denominators 4 and 3) to clear fractions: $$12 \times \left(\frac{1}{4}x - \frac{1}{3}y\right) = 12 \times \frac{7}{4}$$ which simplifies to: $$3x - 4y = 21$$ 4. Now the system is: $$3x - y = -4$$ $$3x - 4y = 21$$ 5. Subtract the first equation from the second to eliminate $3x$: $$\cancel{3x} - 4y - (\cancel{3x} - y) = 21 - (-4)$$ which simplifies to: $$-4y + y = 21 + 4$$ $$-3y = 25$$ 6. Solve for $y$: $$y = \frac{25}{-3} = -\frac{25}{3}$$ 7. Substitute $y = -\frac{25}{3}$ into the first equation: $$3x - \left(-\frac{25}{3}\right) = -4$$ which is: $$3x + \frac{25}{3} = -4$$ 8. Subtract $\frac{25}{3}$ from both sides: $$3x = -4 - \frac{25}{3} = -\frac{12}{3} - \frac{25}{3} = -\frac{37}{3}$$ 9. Divide both sides by 3: $$x = \frac{-\frac{37}{3}}{3} = -\frac{37}{3} \times \frac{1}{3} = -\frac{37}{9}$$ **Final answer:** $$x = -\frac{37}{9}, \quad y = -\frac{25}{3}$$