1. **State the problem:** We have a parallelogram PQRS with segments on its diagonals given as $PQ = x + 29$, $QR = 7x + 5$, $QP = 3y$, and $PS = 8y - 35$. We need to find the values of $x$ and $y$. 2. **Recall the property of parallelograms:** The diagonals of a parallelogram bisect each other. This means the two segments of each diagonal are equal. 3. **Set up equations for $x$:** Since $PQ$ and $QR$ are segments of diagonal $PR$, they must be equal: $$x + 29 = 7x + 5$$ 4. **Solve for $x$:** $$x + 29 = 7x + 5$$ $$x - 7x = 5 - 29$$ $$\cancel{6x} = -24$$ $$-6x = -24$$ $$x = \frac{-24}{-6} = 4$$ 5. **Set up equations for $y$:** Since $QP$ and $PS$ are segments of diagonal $QS$, they must be equal: $$3y = 8y - 35$$ 6. **Solve for $y$:** $$3y = 8y - 35$$ $$3y - 8y = -35$$ $$\cancel{-5y} = -35$$ $$-5y = -35$$ $$y = \frac{-35}{-5} = 7$$ **Final answer:** $$x = 4, \quad y = 7$$