Draw these two lines on a cartesian plane and find their point of intersection: 4x + 2y = 300, 4x + 10y = 340

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

$$(4x + 10y) - (4x + 2y) = 340 - 300$$ $$8y = 40$$ $$y = \frac{40}{8}$$ $$y = 5$$

Using equation (1):

$$4x + 2(5) = 300$$ $$4x + 10 = 300$$ $$4x = 300 - 10$$ $$4x = 290$$ $$x = \frac{290}{4}$$ $$x = 72.5$$

The point of intersection is $(72.5, 5)$.

For line 1: $4x + 2y = 300$
If $x = 0$: $2y = 300 \implies y = 150$. Point: $(0, 150)$.
If $y = 0$: $4x = 300 \implies x = 75$. Point: $(75, 0)$.

For line 2: $4x + 10y = 340$
If $x = 0$: $10y = 340 \implies y = 34$. Point: $(0, 34)$.
If $y = 0$: $4x = 340 \implies x = 85$. Point: $(85, 0)$.

The point of intersection is $\boxed{(72.5, 5)}$.
To draw these lines, plot the calculated points for each line and connect them. The intersection of these two lines will be at $(72.5, 5)$.