How do you solve and graph the inequality #abs(4 – v)< 5#?

Let's start from a graph of a function #y=|v|#.
For non-negative #v# the absolute value of #v# is itself. So, if #v>=0#, #|v|=v# and our function is the same as #y=v#.
For negative #v# the absolute value of #v# is its opposite, #-v#. So, if #v<0#, #|v|=-v# and our function is the same as #y=-v#.
That results in this graph of #y=|v|#:
graph{|x| [-10, 10, -5, 5]}

Now let's draw a graph of #y=|4-v|#. Since #|4-v|=|v-4|#, we will draw a graph of #y=|v-4|#.
According to principles of graph transformation, a graph of #y=|v-4|# is the result of a shift to the right by #4# of a graph of a function #=|v|#. Therefore, our graph looks like this:
graph{|x-4| [-10, 10, -5, 5]}

Now, to solve #|4-v|<5#, we have to find all values of #v#, where the graph lies below the horizontal line that intersects the Y-axis at point #y=5#. Obviously, it's a segment #(-1,9)# because, while #v# is changing from #-1# to #9#, the value of #|4-v|# is changing from #5# down to zero and up to #5#, always staying below the horizontal line #y=5#.
Outside of this segment, that is if #v<=-1# or #y>=9# the value of #|4-v|# is equal or greater than #5#.

For those interested in purely algebraic solution, here is how to do it.
Since, by definition,
#|X|=X# if #X>=0# and
#|X|=-X# if #X<0#,
we will consider two cases.

Case 1. Seeking solutions that satisfy the inequality
#4-v>=0# or, equivalently, #v<=4#.
In this case #|4-v|=4-v# and our original inequality looks like this:
#4-v<5#.
Solution to this is #v > -1#.
Combined with the condition #v<=4# of this case, we have a segment of values of #v#, where #|4-v|<5#:
#-1 < v <= 4#

Case 2. Seeking solutions that satisfy the inequality
#4-v<0# or, equivalently, #v>4#.
In this case #|4-v|=-(4-v)# and our original inequality looks like this:
#-(4-v)<5#, that is #-4+v<5#.
Solution to this is #v < 9#.
Combined with the condition #v>4# of this case, we have another segment of values of #v#, where #|4-v|<5#:
#4 < v < 9#

Now it's appropriate to combine two segments that represent the solutions of an original inequality into one segment since these segments are adjacent:
#-1 < v <= 4# combined with #4 < v < 9# results in a segment #-1 < v < 9#
As you see, we get the same solution as using a graph above (which should not be a surprise).