1. Problem statement: Consider parallelogram QRST with diagonals intersecting at U. 2. Given: the angle at vertex R formed by RU and RS is $32^\circ$. 3. Given: the angle at vertex S formed by SU and SR is $35^\circ$. 4. Given lengths: $RU=3x+3$ and $UT=6$. 5. Key facts and formulas: In any parallelogram the diagonals bisect each other so corresponding diagonal segments are equal; therefore $RU=UT$. 6. Key facts: Opposite sides are parallel, so a diagonal makes equal corresponding angles with parallel sides at opposite vertices; this implies the angle at R formed by RU and RS equals the angle at T formed by UT and TQ, and the angle at S formed by SU and SR equals the angle at Q formed by UQ and QR. 7. Use equality of segments to solve for $x$. 8. Set up the equation: $$3x+3=6$$ 9. Subtract 3 from both sides to isolate the term with $x$. 10. Simplify: $$3x+3-3=6-3$$ 11. Result: $$3x=3$$ 12. Divide both sides by 3 to solve for $x$. 13. Show cancellation when dividing: $$\frac{3x}{3}=\frac{3}{3}$$ 14. Cancellation displayed: $$\frac{\cancel{3}x}{\cancel{3}}=\frac{\cancel{3}}{\cancel{3}}$$ 15. Therefore $x=1$. 16. Find angle $m\angle UTQ$ by corresponding angles: since the angle at R formed by RU and RS is $32^\circ$ and RU is collinear with UT while RS is parallel to TQ, we have $m\angle UTQ=32^\circ$. 17. Find angle $m\angle UQT$ similarly: since the angle at S formed by SU and SR is $35^\circ$ and SU is collinear with UQ while SR is parallel to QR, we have $m\angle UQT=35^\circ$. 18. Optional check: angles in triangle $UTQ$ sum to 180, so $m\angle U + m\angle UTQ + m\angle UQT =180^\circ$. 19. Compute the angle at $U$ inside triangle $UTQ$: $$m\angle T U Q=180- (32+35)$$ 20. Simplify: $$m\angle T U Q=180-67$$ 21. Result: $$m\angle T U Q=113^\circ$$ 22. Final answers: $m\angle UTQ=32^\circ$, $x=1$, $m\angle UQT=35^\circ$.