1. **State the problem:** Given two parallel lines $m \parallel n$ cut by a transversal $t$, find the value of $x$ given the angles $(6x-1)^\circ$ and $(3x+28)^\circ$. 2. **Identify the relationship:** Since $m \parallel n$, the corresponding angles formed by the transversal are equal. Therefore, we set: $$6x - 1 = 3x + 28$$ 3. **Solve the equation:** Subtract $3x$ from both sides: $$6x - 1 - \cancel{3x} = 3x + 28 - \cancel{3x} \implies 3x - 1 = 28$$ Add 1 to both sides: $$3x - 1 + 1 = 28 + 1 \implies 3x = 29$$ Divide both sides by 3: $$\frac{\cancel{3}x}{\cancel{3}} = \frac{29}{3} \implies x = \frac{29}{3}$$ 4. **Final answer:** $$x = \frac{29}{3} \approx 9.67$$