1. **State the problem:** We have quadrilateral NOPQ inscribed in circle R with angles at vertices N, O, P, and Q given as follows: \n- $\angle N = 52^\circ$\n- $\angle O = 74^\circ$\n- $\angle Q = (x - 89)^\circ$\n- $\angle P = (2y - 38)^\circ$\n\nWe need to solve for $x$ and $y$.\n\n2. **Formula and rule:** For any quadrilateral inscribed in a circle, the sum of opposite angles is $180^\circ$. That is, \n$$\angle N + \angle P = 180^\circ$$\nand\n$$\angle O + \angle Q = 180^\circ.$$\n\n3. **Set up equations using the rule:**\n\nFrom $\angle N + \angle P = 180^\circ$:\n$$52 + (2y - 38) = 180$$\n\nFrom $\angle O + \angle Q = 180^\circ$:\n$$74 + (x - 89) = 180$$\n\n4. **Simplify each equation:**\n\nEquation 1:\n$$52 + 2y - 38 = 180$$\n$$\Rightarrow 14 + 2y = 180$$\n\nEquation 2:\n$$74 + x - 89 = 180$$\n$$\Rightarrow x - 15 = 180$$\n\n5. **Solve for $x$:**\n$$x - 15 = 180$$\nAdd 15 to both sides:\n$$x = 180 + 15$$\n$$x = 195$$\n\n6. **Solve for $y$:**\n$$14 + 2y = 180$$\nSubtract 14 from both sides:\n$$2y = 180 - 14$$\n$$2y = 166$$\nDivide both sides by 2 (showing cancellation):\n$$2\cancel{y} = 166 \Rightarrow \cancel{2}y = \frac{166}{2}$$\n$$y = 83$$\n\n**Final answers:**\n$$x = 195$$\n$$y = 83$$