In the diagram above RT and PS are parallel <RPQ = 70°, <PQR = 2x, <RQS = 3x + 15° and <PRQ = y°. Find the values of x and y. Factorize the expression 2x + xy + 6 + 3y

Angles $\angle PQR$ and $\angle RQS$ are angles on a straight line $PS$, so their sum is $180^\circ$.

$$\angle PQR + \angle RQS = 180^\circ$$

Substitute the given expressions for the angles:

$$2x + (3x + 15^\circ) = 180^\circ$$

Combine like terms:

$$5x + 15^\circ = 180^\circ$$

Subtract $15^\circ$ from both sides:

$$5x = 180^\circ - 15^\circ$$ $$5x = 165^\circ$$

Divide by 5:

$$x = \frac{165}{5}$$ $$x = 33$$

First, calculate $\angle PQR$ using the value of $x$:

$$\angle PQR = 2x = 2(33^\circ) = 66^\circ$$

The sum of angles in a triangle is $180^\circ$. For $\triangle PQR$:

$$\angle RPQ + \angle PQR + \angle PRQ = 180^\circ$$

Substitute the known angle values:

$$70^\circ + 66^\circ + y^\circ = 180^\circ$$

Add the known angles:

$$136^\circ + y^\circ = 180^\circ$$

Subtract $136^\circ$ from both sides:

$$y^\circ = 180^\circ - 136^\circ$$ $$y = 44$$

The values are $\boxed{x = 33, y = 44}$.

Group the first two terms and the last two terms:

$$(2x + xy) + (6 + 3y)$$

From the first group, factor out $x$:

$$x(2 + y)$$

From the second group, factor out 3:

$$3(2 + y)$$

Now the expression is:

$$x(2 + y) + 3(2 + y)$$

$$(2 + y)(x + 3)$$

The factored expression is $\boxed{(x+3)(y+2)}$.