There are two ways to solve this:
Method 1. You can add #d# to both sides, since you are subtracting #d#. This is pretty quick, but it can be confusing because normally you add, subtract, etc. numbers, not variables.
Method 2. You can subtract #38# from both sides (because #38-d=-d+38#), then divide by #-1#. This takes a while, plus you have to flip the inequality symbol, turning the #># into a #<#. However, you only have to use numbers--you don't add, subtract, etc. the variable.
Both methods give the same answer; it's a matter of preference. I will do both.
Method 1:
#38-d<37#
Add #d# to both sides.
#38<37+d#
Subtract 37 from both sides (see footnote):
#38-37>d#
Simplify:
#1>d#, which can be rewritten as #d>1#.
Therefore, #d>1#.
Method 2:
#38-d<37#
Subtract #38# from both sides:
-d<-1
Divide by #-1# and flip the #<#, making it a #># (you can see why: #1<2#, divide by #-1# and you get #-1<-2# -- whoops! Turn the #<# into a #># and everything's fine.):
d>1
Therefore, #d>1#.
Let's check. Since #3>1#, we can substitute #3# for #d#:
#38-3<37#
#35<37#
We're good!
So, the solution is #d>1.#
Footnote: #37+d=d+37# by the Commutative Property of Addition, so you subtract to cancel out the addition. (I put this separately to stop the paragraph from being too long.)
