Solve the system of linear equations: 2x + 2y = 10 and 3x − y = 3.

We have the following system:

$$2x + 2y = 10 \quad (1)$$ $$3x - y = 3 \quad (2)$$

Multiplying equation (2) by 2 gives:

$$2(3x - y) = 2(3)$$ $$6x - 2y = 6 \quad (3)$$

$$(2x + 2y) + (6x - 2y) = 10 + 6$$ $$2x + 6x + 2y - 2y = 16$$ $$8x = 16$$

Divide both sides by 8:

$$\frac{8x}{8} = \frac{16}{8}$$ $$x = 2$$

Using equation (2):

$$3x - y = 3$$

Substitute $x = 2$:

$$3(2) - y = 3$$ $$6 - y = 3$$

Subtract 6 from both sides:

$$-y = 3 - 6$$ $$-y = -3$$

Multiply both sides by -1:

$$y = 3$$

The solution to the system of equations is:

$$x = \boxed{2}$$ $$y = \boxed{3}$$