1. **Problem statement:** Given parallel lines TY || HT and transversals BG and KP, with angles $m\angle KXY = (15x - 2)^\circ$, $m\angle TWP = (7x + 6)^\circ$, $m\angle MRH = (6y + 4)^\circ$, and $m\angle FMR = (9y - 11.5)^\circ$, find $x$, $y$, and the measures of angles $KXY$, $TWP$, $MRH$, $FMR$, $BMF$, and $XWT$. 2. **Identify relationships:** Since TY || HT and BG, KP are transversals, corresponding or alternate interior angles are equal. 3. **Set up equations:** - $m\angle KXY = m\angle TWP$ because they are corresponding angles. So, $$15x - 2 = 7x + 6$$ 4. **Solve for $x$:** $$15x - 2 = 7x + 6$$ $$15x - \cancel{2} - 7x = \cancel{6} + 6$$ $$8x - 2 = 6$$ $$8x = 6 + 2$$ $$8x = 8$$ $$x = \frac{8}{8} = 1$$ 5. **Use $x=1$ to find angle measures:** - $m\angle KXY = 15(1) - 2 = 15 - 2 = 13^\circ$ - $m\angle TWP = 7(1) + 6 = 7 + 6 = 13^\circ$ 6. **Use $m\angle MRH = m\angle FMR$ because they are vertical angles:** $$6y + 4 = 9y - 11.5$$ 7. **Solve for $y$:** $$6y + 4 = 9y - 11.5$$ $$6y + 4 - 9y = 9y - 11.5 - 9y$$ $$-3y + 4 = -11.5$$ $$-3y = -11.5 - 4$$ $$-3y = -15.5$$ $$y = \frac{-15.5}{-3} = \frac{15.5}{3} = 5.1667$$ 8. **Calculate angles $MRH$ and $FMR$:** - $m\angle MRH = 6(5.1667) + 4 = 31 + 4 = 35^\circ$ (rounded) - $m\angle FMR = 9(5.1667) - 11.5 = 46.5 - 11.5 = 35^\circ$ 9. **Find $m\angle BMF$ and $m\angle XWT$:** - Since $BMF$ and $FMR$ are supplementary (linear pair), $$m\angle BMF = 180^\circ - m\angle FMR = 180 - 35 = 145^\circ$$ - Since $XWT$ and $TWP$ are supplementary, $$m\angle XWT = 180^\circ - m\angle TWP = 180 - 13 = 167^\circ$$ **Final answers:** \[x = 1,\quad y = 5.1667,\quad m\angle KXY = 13^\circ,\quad m\angle TWP = 13^\circ,\quad m\angle MRH = 35^\circ,\quad m\angle FMR = 35^\circ,\quad m\angle BMF = 145^\circ,\quad m\angle XWT = 167^\circ\]