Solve the system of linear equations: x + 2y = 6 and 3x + 2y = 14.

We have the following system:

$$x + 2y = 6 \quad (1)$$ $$3x + 2y = 14 \quad (2)$$

Subtract equation (1) from equation (2) to eliminate $y$:

$$(3x + 2y) - (x + 2y) = 14 - 6$$ $$3x + 2y - x - 2y = 8$$ $$2x = 8$$

Divide both sides by 2:

$$\frac{2x}{2} = \frac{8}{2}$$ $$x = 4$$

Using equation (1):

$$x + 2y = 6$$

Substitute $x = 4$:

$$4 + 2y = 6$$

Subtract 4 from both sides:

$$2y = 6 - 4$$ $$2y = 2$$

Divide both sides by 2:

$$\frac{2y}{2} = \frac{2}{2}$$ $$y = 1$$

The solution to the system of equations is:

$$x = \boxed{4}$$ $$y = \boxed{1}$$