1. **Problem statement:** Given two parallel lines $m \parallel n$ and a transversal $t$ intersecting them, find the value of $x$ given the angles $(2x+5)^\circ$ and $(9x+10)^\circ$ formed at the intersections. 2. **Key concept:** When a transversal intersects two parallel lines, corresponding angles are equal. 3. **Set up the equation:** Since $(2x+5)^\circ$ and $(9x+10)^\circ$ are corresponding angles, we have: $$2x + 5 = 9x + 10$$ 4. **Solve for $x$:** $$2x + 5 = 9x + 10$$ Subtract $2x$ from both sides: $$\cancel{2x} + 5 = \cancel{2x} + 9x + 10 \implies 5 = 7x + 10$$ Subtract 10 from both sides: $$5 - 10 = 7x + 10 - 10 \implies -5 = 7x$$ Divide both sides by 7: $$\frac{-5}{\cancel{7}} = \frac{7x}{\cancel{7}} \implies x = -\frac{5}{7}$$ 5. **Final answer:** $$x = -\frac{5}{7}$$