At the point $(x_0, y_0)$ where they intersect, $(x_0, y_0)$ satisfies both equations, that is:
$$
\left\{\begin{array}{rcl}y_0 = 2x_0 + 4 \\ y_0 = 3x_0 + 5 \end{array}\right.
$$
In particular, both of the right-hand sides are equal to $y_0$, so they are equal to one another:
$$2x_0 + 4 = 3x_0 +5.$$
Now, rearranging gives
$$x_0 = -1.$$
Again, the point $(x_0, y_0)$ is on both lines, so we can substitute it in either equation, say, the first, and get a value for $y_0$:
$$y_0 = 2(-1) + 4 = 2.$$
So, the intersection point is
$$(x_0, y_0) = (-1, 2),$$
and we can check this, if we like, by substituting in the other equation.
It may be an instructive exercise, by the way, to work this out for a general pair of lines in the plane specified this way, that is, for the lines
$$
\left\{\begin{array}{rcl}y &=& m\phantom{'} x + b\phantom{'} \\ y &=& m' x + b' \end{array}\right.
$$
Note that when $m = m'$ there are either no solutions or infinitely many solutions---what do these special cases correspond to geometrically?