1. **Problem statement:** Given two parallel lines $a \parallel b$ and a transversal, with an angle of $120^\circ$ at the upper intersection and expressions $7x - 10$ and $10y - 10$ at the lower intersection, find $x + y$. 2. **Key fact:** When two lines are parallel, alternate interior angles are equal, and corresponding angles are equal. 3. The angle adjacent to the $120^\circ$ angle on the upper intersection is $180^\circ - 120^\circ = 60^\circ$ because they form a linear pair. 4. Since $a \parallel b$, the angle corresponding to $7x - 10$ on the lower intersection equals $60^\circ$. 5. Similarly, the angle corresponding to $10y - 10$ on the lower intersection is supplementary to $7x - 10$, so their sum is $180^\circ$: $$ (7x - 10) + (10y - 10) = 180 $$ 6. From step 4, set $7x - 10 = 60$: $$ 7x - 10 = 60 $$ Add 10 to both sides: $$ 7x = 60 + 10 $$ $$ 7x = 70 $$ Divide both sides by 7: $$ x = \frac{\cancel{7}0}{\cancel{7}} = 10 $$ 7. Substitute $x=10$ into the supplementary angle equation: $$ (7(10) - 10) + (10y - 10) = 180 $$ Simplify: $$ (70 - 10) + (10y - 10) = 180 $$ $$ 60 + 10y - 10 = 180 $$ $$ 10y + 50 = 180 $$ Subtract 50 from both sides: $$ 10y = 180 - 50 $$ $$ 10y = 130 $$ Divide both sides by 10: $$ y = \frac{\cancel{1}30}{\cancel{1}0} = 13 $$ 8. Finally, find $x + y$: $$ x + y = 10 + 13 = 23 $$ **Answer:** $x + y = 23$