1. **Problem statement:** Given quadrilateral with vertices U, T, X, V and point W on segment VX, where UV is parallel to TW. Given lengths: $UT=12$, $TX=24$, $UV=22$, and we need to find $VX$. 2. **Key concept:** Since $UV \parallel TW$, triangles $UVT$ and $T WX$ are similar by the AA criterion (corresponding angles are equal). 3. **Set up ratio from similarity:** Corresponding sides are proportional: $$\frac{UV}{TW} = \frac{UT}{TX} = \frac{VT}{WX}$$ 4. We know $UV=22$, $UT=12$, $TX=24$. So, $$\frac{22}{TW} = \frac{12}{24} = \frac{1}{2}$$ 5. Solve for $TW$: $$\frac{22}{TW} = \frac{1}{2} \implies 22 \times 2 = TW \implies TW = 44$$ 6. Since $W$ lies on $VX$, and $TW=44$, $VX = VW + WX = TW + WX$ (because $W$ is on $VX$ between $V$ and $X$). 7. From similarity, $\frac{VT}{WX} = \frac{1}{2}$, but $VT$ is unknown. However, since $W$ lies on $VX$, and $TW=44$, the length $VX$ equals $TW$ plus $WX$. 8. But $TW$ is $44$, and $TX=24$, so $WX = TX - TW = 24 - 44 = -20$, which is impossible. So $W$ must be between $V$ and $X$ such that $VX = TW = 44$. 9. Therefore, the length $VX$ is $44$ units. **Final answer:** $$VX = 44$$