1. **State the problem:** We are given three angles related by a transversal intersecting two parallel lines: angles are $(8x - 10)^\circ$, $(6y + 20)^\circ$, and $(7x)^\circ$. We need to find the values of $x$ and $y$. 2. **Identify angle relationships:** Since $(8x - 10)^\circ$ and $(6y + 20)^\circ$ are on opposite sides of the transversal and between the parallel lines, they are alternate interior angles and thus equal: $$8x - 10 = 6y + 20$$ 3. **Adjacent angles on the same side of the transversal:** The angles $(6y + 20)^\circ$ and $(7x)^\circ$ are adjacent and on the same side of the transversal, so they are supplementary (sum to $180^\circ$): $$6y + 20 + 7x = 180$$ 4. **Write the system of equations:** $$\begin{cases} 8x - 10 = 6y + 20 \\ 6y + 20 + 7x = 180 \end{cases}$$ 5. **Simplify the first equation:** $$8x - 10 = 6y + 20 \implies 8x - 6y = 30$$ 6. **Simplify the second equation:** $$6y + 7x + 20 = 180 \implies 7x + 6y = 160$$ 7. **Add the two equations:** $$8x - 6y = 30$$ $$7x + 6y = 160$$ Adding: $$8x - 6y + 7x + 6y = 30 + 160 \implies 15x = 190$$ 8. **Solve for $x$:** $$x = \frac{190}{15} = \frac{38}{3} \approx 12.67$$ 9. **Substitute $x$ into one equation to find $y$:** Use $8x - 6y = 30$: $$8 \times \frac{38}{3} - 6y = 30$$ $$\frac{304}{3} - 6y = 30$$ $$-6y = 30 - \frac{304}{3} = \frac{90}{3} - \frac{304}{3} = -\frac{214}{3}$$ $$y = \frac{214}{18} = \frac{107}{9} \approx 11.89$$ **Final answers:** $$x = \frac{38}{3} \approx 12.67, \quad y = \frac{107}{9} \approx 11.89$$