1. **State the problem:** We are given expressions for segments SU and SW in terms of $b$ and need to find the value of $b$ that makes quadrilateral TUVW a parallelogram. 2. **Recall the property of parallelograms:** The diagonals of a parallelogram bisect each other. This means the segments formed by the intersection point S on the diagonals are equal: specifically, $SU = SW$. 3. **Set up the equation:** Given $$SU = 2b + 22$$ $$SW = 4b + 20$$ Since $SU = SW$ for a parallelogram, we have $$2b + 22 = 4b + 20$$ 4. **Solve for $b$:** Subtract $2b$ from both sides: $$\cancel{2b} + 22 = \cancel{2b} + 4b + 20 \Rightarrow 22 = 2b + 20$$ Subtract 20 from both sides: $$22 - 20 = 2b + \cancel{20} - \cancel{20} \Rightarrow 2 = 2b$$ Divide both sides by 2: $$\frac{2}{\cancel{2}} = \frac{2b}{\cancel{2}} \Rightarrow 1 = b$$ 5. **Conclusion:** The value of $b$ that makes TUVW a parallelogram is $$\boxed{1}$$