1. **State the problem:** Solve the simultaneous equations: a) $y = 4x - 10$ b) $y + 2x = 14$ 2. **Use substitution method:** Since $y$ is expressed in terms of $x$ in equation (a), substitute $y = 4x - 10$ into equation (b): $$4x - 10 + 2x = 14$$ 3. **Simplify the equation:** $$4x + 2x - 10 = 14$$ $$6x - 10 = 14$$ 4. **Isolate $x$:** $$6x - 10 + 10 = 14 + 10$$ $$6x = 24$$ 5. **Solve for $x$:** $$\frac{\cancel{6}x}{\cancel{6}} = \frac{24}{6}$$ $$x = 4$$ 6. **Find $y$ using $x=4$ in equation (a):** $$y = 4(4) - 10$$ $$y = 16 - 10$$ $$y = 6$$ 7. **Check the solution in equation (b):** $$y + 2x = 6 + 2(4) = 6 + 8 = 14$$ The solution satisfies both equations. **Final answer:** $$x = 4, \quad y = 6$$