1. **Problem statement:** We have two parallel lines $q$ and $r$ cut by a transversal $s$. The angles formed at the intersections are given as $(7x + 10)^\circ$ on line $q$ and $(3x + 20)^\circ$ on line $r$. We need to find the value of $x$. 2. **Key concept:** When two parallel lines are cut by a transversal, corresponding angles are equal. 3. **Set up the equation:** Since the angles are corresponding, we have: $$7x + 10 = 3x + 20$$ 4. **Solve for $x$:** Subtract $3x$ from both sides: $$7x + 10 - \cancel{3x} = \cancel{3x} + 20 - 3x$$ $$4x + 10 = 20$$ Subtract 10 from both sides: $$4x + 10 - 10 = 20 - 10$$ $$4x = 10$$ Divide both sides by 4: $$\frac{4x}{\cancel{4}} = \frac{10}{\cancel{4}}$$ $$x = \frac{10}{4} = 2.5$$ 5. **Check the options:** The options are 15, 10, 11, 14. None match $2.5$ exactly, so check if the angles are alternate interior angles or supplementary. 6. **Alternate interior angles:** If the angles are alternate interior angles, they are equal, so $x=2.5$. 7. **If angles are supplementary:** Since the transversal crosses parallel lines, the angles on the same side of the transversal are supplementary: $$ (7x + 10) + (3x + 20) = 180 $$ $$ 10x + 30 = 180 $$ $$ 10x = 150 $$ $$ x = 15 $$ 8. **Conclusion:** The correct interpretation is that the angles are supplementary, so $x=15$. **Final answer:** $x=15$