1. **State the problem:** Given a parallelogram WXYZ with sides labeled as follows: - WX = $b + 11$ - XY = $3a + 7$ - YZ = $2b$ - WZ = $4a$ We need to find the values of $a$ and $b$, then find the sum $a + b$. 2. **Recall the property of parallelograms:** Opposite sides of a parallelogram are equal in length. So, we have: $$WX = YZ$$ $$XY = WZ$$ 3. **Set up equations:** From $WX = YZ$: $$b + 11 = 2b$$ From $XY = WZ$: $$3a + 7 = 4a$$ 4. **Solve for $b$:** $$b + 11 = 2b$$ Subtract $b$ from both sides: $$11 = 2b - b$$ $$11 = b$$ 5. **Solve for $a$:** $$3a + 7 = 4a$$ Subtract $3a$ from both sides: $$7 = 4a - 3a$$ $$7 = a$$ 6. **Find the sum $a + b$:** $$a + b = 7 + 11 = 18$$ **Final answer:** $$a = 7, \quad b = 11, \quad a + b = 18$$