How can I solve a rate problem given the rate and time in terms $r + 200$ and $t - 2$ The Kaplan solution to the problem just says backsolve from the answer choices till values are found that satisfy the conditions. This is less than ideal if I'm trying to figure out how to setup the actual math.
The question:

A bike messenger pedals at a constant rate from Location A to Location B, a distance of 4,800 meters. If she had traveled at a constant rate that was 200 meters per minute faster than her actual rate, she would have arrived at Location B 2 minutes earlier. How many minutes did the bike messenger spend travelling from Location A to Location B.

I set up the time like so:
$$ \begin{align} \frac{4,800}{r + 200} + 2 = t \end{align}$$
And then solved for $4,800 = (r + 200) (\frac{4,800}{r + 200})$, but that doesn't lead anywhere.
How would I correctly solve this problem?
 A: Your $t$ should be $\frac {4800}r$, which is the time the messenger actually took.  That gives you one equation in one unknown.
A: We have $t=\frac{4800}r$,
$$\frac{4800}{r}-\frac{4800}{r+200}=2$$
which give us a quadratic equation, solve for $r$, then compute $\frac{4800}r$ to get the time taken.
Edit:
View $200$ as $1$ distance unit. Then the problem would be 

A bike messenger pedals at a constant rate from Location A to Location B, a distance of $24$ distance unit. If she had traveled at a constant rate that was $1$ distance unit per minute faster than her actual rate, she would have arrived at Location B $2$ minutes earlier. How many minutes did the bike messenger spend travelling from Location A to Location B.

$$\frac{24}r-\frac{24}{r+1}=2$$
$$\frac{12}r-\frac{12}{r+1}=1$$
$$12=r(r+1)$$
$$r^2+r-12=0$$
$$(r+4)(r-3)=0$$
$$r=3$$
$$t=\frac{24}{3}=8$$
A: Let the original speed be $s$ (m/min) and the original time taken be $t$ (min). Since distance is speed $\times$ time and the distance covered is constant, $st = (s + 200)(t - 2) = 4800$, or:
$$st = st - 2s + 200t - 400$$
$$\Rightarrow 2s = 200t - 400, \ s = 100t-200$$
and since $st = 4800$, plug in $s$ to get $(100t - 200)t = 4800 \Rightarrow (t - 2)t = 48$. If $t$ doesn't jump out immediately, rewrite this as the quadratic $t^2 - 2t - 48 = 0$, giving $(t - 8)(t + 6) = 0$ or $t = 8$.
