Set up your equations:
C = # of caviar
B = # bologna
L = # of liverwurst
C + B + L = 100
5C + 2B + L/10 = 100
multiply the 2nd equation by 10 and subtract the two
49C + 19B = 900
Now, there are lots of ways to do that, but you need both C and B to be integers. Notice that 900 is divisible by 10. Divide the whole equation by 10
$4.9 C + $1.9 B = $90
Now, we know she need to make exactly $90 selling just caviar and bologna. Each time she sells a sandwich, she'll have to give a dime in change. To get back to an exact dollar figure, she'll need to sell some multiple of 10 sandwiches. So, make the substitution
C + B = 10 n
Now, solve for C and substitute
49 n - 3 B = 90
Remembering that n and B are both integers, n must be a multiple of 3 (divide that equation by 3 and see). So try, n=3, 6, 9, etc.
For n=3, B = 19
For n=6, B = 68
For n=9, B = 117 (which is impossible.)
If B = 19, then C = 11 and L = 70.
If B = 68, then C = -8 which is impossible.
So, the only possible answer is 11 caviar sandwiches, 19 bologna sandwiches and 70 Liverwurst sandwiches.
...What are you, a 1st grader?
This is impossible to answer because there can be an infinite number of solutions to this problem.
A possible answer can be:
18 Caviar Sandiwiches = $90
4 Bologna Sandwiches = $8
20 Liverwurst Sandwiches = $2
This all equals $100. There can be many more solutions to this problem but this is one of them.
You example only equals to 42 sandwiches... might want to check the question again.