1. **State the problem:** We have the system of linear equations: $$\frac{1}{3}x + \frac{1}{6}y = 7$$ $$ax - 4y = 10$$ We need to find the value of $a$ such that the system has no solution. 2. **Recall the condition for no solution:** Two lines have no solution if they are parallel but not the same line. Parallel lines have equal slopes but different intercepts. 3. **Find the slope of the first line:** Rewrite the first equation in slope-intercept form $y = mx + b$: $$\frac{1}{3}x + \frac{1}{6}y = 7$$ Multiply both sides by 6 to clear denominators: $$6 \times \left(\frac{1}{3}x + \frac{1}{6}y\right) = 6 \times 7$$ $$2x + y = 42$$ Isolate $y$: $$y = -2x + 42$$ So the slope of the first line is $m_1 = -2$. 4. **Find the slope of the second line:** Rewrite the second equation in slope-intercept form: $$ax - 4y = 10$$ Isolate $y$: $$-4y = -ax + 10$$ $$y = \frac{a}{4}x - \frac{10}{4}$$ So the slope of the second line is $m_2 = \frac{a}{4}$. 5. **Set slopes equal for parallel lines:** $$m_1 = m_2$$ $$-2 = \frac{a}{4}$$ Multiply both sides by 4: $$4 \times (-2) = 4 \times \frac{a}{4}$$ $$-8 = \cancel{4} \times \frac{a}{\cancel{4}}$$ So, $$a = -8$$ 6. **Conclusion:** The value of $a$ that makes the system have no solution (parallel lines) is $\boxed{-8}$.