1. **State the problem:** We are given two equations: $$\frac{4a + 2}{11} = 2$$ and $$3b - 2 = a - 19$$ We need to find the value of $b$. 2. **Solve the first equation for $a$:** Multiply both sides by 11 to eliminate the denominator: $$4a + 2 = 2 \times 11$$ $$4a + 2 = 22$$ Subtract 2 from both sides: $$4a = 22 - 2$$ $$4a = 20$$ Divide both sides by 4: $$a = \frac{20}{4} = 5$$ 3. **Substitute $a = 5$ into the second equation:** $$3b - 2 = 5 - 19$$ Simplify the right side: $$3b - 2 = -14$$ Add 2 to both sides: $$3b = -14 + 2$$ $$3b = -12$$ Divide both sides by 3: $$b = \frac{-12}{3} = -4$$ **Final answer:** $$b = -4$$