A bus traveling between $2$ towns arrives $7$ minutes late when traveling $40$ mph. When traveling $36$ mph, the bus arrives $22$ minutes late. What is the distance between the two towns?
Here's what I did:
$x=$ original time
$40(x+7) = 36(x+22)$
$x=128$, so when you substitute $x$ for one of the expressions, you get $40(128+7)=5400$, so 5,400 miles?
Your methodology is generally correct. However, note that if you are considering: $$40(x+\color{red}{7})=36(x+\color{red}{22})$$ Then you are considering the delay to be $7$ and $22$ hours, not minutes. Therefore, you must solve the following: $$40\left(x+\frac{7}{60}\right)=36\left(x+\frac{22}{60}\right)$$ Giving the answer for $x$ in hours. Then, you can do the same as you've done before (Substitute $x$ into one of the expressions).
Your problem is that the way you set things up, $x$ is a number of minutes (which you can see when you notice that we are adding $7$ or $22$ minutes to $x$). So you correctly deduced that $x$ is $128$ minutes and then went astray by multiplying a number of miles per hour times a number of minutes.
What you should have done in the last step is to multiply $\frac{40}{60}$ miles per minute by $128+7$ minutes, getting an answer of $90$ miles.
First note that $40$ mph$=\frac 23 $m/min and $36$ mph $=\frac 35$ m/min.
Let $D$=distance between towns, and $T$ =time taken for journey to get to destination on time.
Using distance [m]= speed [m/min] $\times$ time [min], we have
$$D=\frac 23 (T+7)=\frac 35 (T+22)\\ D=\frac {\overbrace{2T+14}^P}{\underbrace{\;\;\;3\;\;\;}_R}=\frac {\overbrace{3T+66}^Q}{\underbrace{\;\;\;5\;\;\;}_S}=\frac {\overbrace{132-42}^{2Q-3P}}{\underbrace{10-9}_{2S-3R}}=\color{red}{90}$$ using linear componendo and dividendo.