1. **State the problem:** Solve the system of linear equations: $$x + 3y = 22$$ $$x = 4$$ 2. **Use substitution:** Since $x = 4$, substitute this value into the first equation: $$4 + 3y = 22$$ 3. **Isolate $y$:** Subtract 4 from both sides: $$\cancel{4} + 3y = 22 - \cancel{4}$$ $$3y = 18$$ 4. **Solve for $y$:** Divide both sides by 3: $$\frac{3y}{\cancel{3}} = \frac{18}{\cancel{3}}$$ $$y = 6$$ 5. **Final answer:** The solution to the system is: $$x = 4, \quad y = 6$$