I am literally trying to repair a fence in my backyard right now. There are 4 fence posts in a line. The distance between post 1 and post 2 is 67". The distance between post 2 and post 3 is 88". The distance between post 3 and post 4 is 89.5"
I have fence panels that are 5.375" wide.
I am trying to find a gap distance such that the gap between each post and its adjacent fence panels and the gap between each fence panel is the same between all 3 sections of the fence.
I think I have come up with 3 equations, but I don't know how to solve it, which is why I am turning to this community to help. So, this is what I have so far:
Let `b` = # of boards between posts
let `g` = gap between.
So, for the first section:
67" = (b*5.375) + ((b + 1)*g)
And the second and third section is the equivalent equation, except 67 is 88 and 89.5 respectively.
Any way I can figure out the # of boards and the gap distance for these 3 sections mathematically?
Maybe I'm overthinking things...Is there an easier way to determine an equivalent gap distance?
Edit: And here is the completed fence, as promised. First section gap: 31mm, second section 33mm, third section 35mm :)

Let
We then (as you correctly found) have $$L_i=n_ia+(n_i+1)b \implies n_i=\frac{L_i-b}{a+b}.$$
We demand that $n_i$ is some integer for the given $L_i$ and $a$ and with the variable $b$, but there is no guarantee that this will be solvable in all three cases ($i=1,2,3$) - in fact, it is very unlikely.
If you define some $\epsilon$ that $b$ is allowed to change for each panel, then there is some $b\pm\epsilon$ that should work (it is all a matter of how small we can make $\epsilon$), but since this is a practical problem, I'd go with a guesstimate and then simply do trial-and-error.
EDIT:
Here's a plot for $i=1$:
Here $b$ is on the $x$-axis, plotted in the range between 0" and 2". The values of $b$ that give integer values (i.e. number of panels) $8,9,10,11,12$ are
$$2.66667,1.8625,1.20455,0.65625,0.192308.$$
I thought values of $b>3$ would be unacceptable, so I left those solutions out. Now, let's see if there are any of these $b$ that crop up again for the other $i$s:
For $i=2$ we have (for number of fence panels $11,12,13,14,15,16$):
$$2.40625,1.80769,1.29464,0.85,0.460938,0.117647.$$
And for $i=3$ we have (for number of fence panels $11,12,13,14,15,16$):
$$2.53125,1.92308,1.40179,0.95,0.554688,0.205882.$$
Now all one has to do is to find the triplet that lies in the smallest interval.
We can plot these values of $b$ to easily see which ones are candidates:
$i=1$ is represented by the blue dots. From this it can be seen that the best approximation of $b$ is given by the lowest values of $b$ in each series (which is nice, because then there won't be such large gaps between the panels).
If a gap of $\sim 0.167$ is too small, the next best thing would be to choose 11, 15 and 15 panels, given a gap of $\sim 0.56$.