1. **State the problem:** We have a cyclic quadrilateral ABCD inscribed in a circle with interior angles at vertices A, B, C, and D given as $26y^\circ$, $3x^\circ$, $2x^\circ$, and $21y^\circ$ respectively. 2. **Recall the property of cyclic quadrilaterals:** The sum of the interior angles of any quadrilateral is $360^\circ$. Also, opposite angles of a cyclic quadrilateral sum to $180^\circ$. 3. **Set up equations using opposite angles:** Angle A and angle C are opposite, so: $$26y + 2x = 180$$ Angle B and angle D are opposite, so: $$3x + 21y = 180$$ 4. **Solve the system of equations:** From the first equation: $$26y = 180 - 2x$$ From the second equation: $$3x + 21y = 180$$ Substitute $y$ from the first into the second: $$y = \frac{180 - 2x}{26}$$ So, $$3x + 21 \times \frac{180 - 2x}{26} = 180$$ Multiply both sides by 26 to clear denominator: $$26 \times 3x + 21(180 - 2x) = 26 \times 180$$ $$78x + 3780 - 42x = 4680$$ Simplify: $$36x + 3780 = 4680$$ $$36x = 4680 - 3780$$ $$36x = 900$$ $$x = \frac{900}{36} = 25$$ 5. **Find $y$:** $$26y = 180 - 2(25) = 180 - 50 = 130$$ $$y = \frac{130}{26} = 5$$ 6. **Calculate each interior angle:** $$\angle A = 26y = 26 \times 5 = 130^\circ$$ $$\angle B = 3x = 3 \times 25 = 75^\circ$$ $$\angle C = 2x = 2 \times 25 = 50^\circ$$ $$\angle D = 21y = 21 \times 5 = 105^\circ$$ 7. **Check sum:** $$130 + 75 + 50 + 105 = 360^\circ$$ **Final answer:** $x = 25$, $y = 5$, and the interior angles are $130^\circ$, $75^\circ$, $50^\circ$, and $105^\circ$ respectively.