Question 1062817
Let's replace r with something you'd find in the set Z. The set Z stands for the set of integers. An easy integer to deal with is r = 0


5r+2 = 5*0+2 = 2


So 2 is in the set C. If C was a subset of D, then 2 must also be in set D for it to work. Let's find out. Set 10s-3 equal to 2 and solve for s


10s-3 = 2
10s = 5
s = 5/10
s = 1/2
s = 0.5 <--- this is NOT an integer


since s is NOT an integer, this means that 2 is NOT in set D. So set C is definitely not a subset of D


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Let's repeat the steps above, but now start with set D. Plug in s = 0


10s-3 = 10*0 - 3 = -3


So -3 is definitely in set D. Is it in set C? Set -3 equal to 5r+2 and solve for r to find out


5r+2 = -3
5r = -5
r = -1


If r = -1, then it leads to -3. This tells us that -3 is definitely in set C


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We can repeat the last section with other values of s to find corresponding values of r.


It turns out that with a bit of trial and error, we can manipulate the form for set D to match what we see in set C, like so...


10s - 3
2*5s - 3
2*5s + 2 - 5
2*5s - 5 + 2
5*2s - 5 + 2
5*(2s - 1) + 2


The (2s-1) portion is an integer so we basically have this format 5*(integer) + 2 which is what set C is saying.


Therefore, any number in set D is also in set C.


This means set D is a subset of set C.