Parameterize the line through P=(−3,−1) and Q=(0,7) so that the points P and Q correspond to the parameter values t=13 and 16.
I have found the direction vector <3,8> and then have plugged in to find the following:
r=<-3,-1> +t<3,8>
x=3t-3 y=8t-1
I am lost as to the next steps to scale this appropriately to come up with the correct equation.
Here is an alternate approach beginning with what you started.
$$r=\langle -3,-1\rangle+t\langle 3,8 \rangle$$
Now this takes us from the first to the second point too quickly, in one unit of $t$ when it should take three units of $t$. So we replace $t$ with $t/3$ to get
$$r=\langle -3,-1\rangle+\frac{t}{3}\langle 3,8 \rangle$$
$$r=\langle -3,-1\rangle+t\left\langle 1,\frac{8}{3} \right\rangle$$
But now, we are arriving at the second point when $t=3$ which is $13$ units of $t$ too soon. So we should delay by $13$ units of $t$ so we arrive when $t=16$.
$$r=\langle -3,-1\rangle+(t-13)\left\langle 1,\frac{8}{3} \right\rangle$$
$$r=\langle -3,-1\rangle+t\left\langle 1,\frac{8}{3} \right\rangle-13\left\langle 1,\frac{8}{3} \right\rangle$$
$$r=\left\langle -16,-\frac{107}{3}\right\rangle+t\left\langle 1,\frac{8}{3} \right\rangle$$
ADDENDUM:
Note that all of this may be summarized in a formula.
If for $t_0<t_1$, $r(t)$ is a linear function satisfying $r(t_0)=\langle x_0,y_0\rangle$ and $r(t_1)=\langle x_1,y_1\rangle$ and if $t_0\le t\le t_1$ then
$$ r(t)=\langle x_0,y_0\rangle+\frac{t-t_0}{t_1-t_0}\langle x_1-x_0,y_1-y_0\rangle $$