1. **Problem Statement:** Given that lines KL and MN are parallel, and lines LM and NO are parallel, find the value of $x$ given the angles $3x^\circ$, $96^\circ$, and $2y^\circ$ in the adjacent triangles sharing point $M$. 2. **Identify the relationships:** Since $KL \parallel MN$ and $LM \parallel NO$, the angles formed at point $M$ have special relationships due to parallel lines and transversal properties. 3. **Use the fact that the sum of angles around point $M$ is $360^\circ$:** The angles at $M$ are $3x^\circ$, $96^\circ$, and $2y^\circ$, plus the angle adjacent to $2y^\circ$ which is supplementary to $2y^\circ$ because $LM \parallel NO$. 4. **Use alternate interior angles:** Because $KL \parallel MN$, angle $3x^\circ$ and angle $2y^\circ$ are corresponding or alternate interior angles, so they are equal: $$3x = 2y$$ 5. **Sum of angles on a straight line:** The angle $96^\circ$ and the angle adjacent to it (which is $2y^\circ$) form a straight line, so: $$96 + 2y = 180$$ 6. **Solve for $y$:** $$2y = 180 - 96 = 84$$ $$y = \frac{84}{2} = 42$$ 7. **Substitute $y$ back into $3x = 2y$:** $$3x = 2 \times 42 = 84$$ $$x = \frac{84}{3} = 28$$ **Final answer:** $$\boxed{28}$$