Question 851508
Let the numbers be A, B, C, D
Here's what we are told:
B = 3A-9 (the second is 9 less than thrice the first)
C = A+B+6 (the third is six more than the sum of the first two numbers)
A + B + C + D = 190 (the sum of number is 190) <<--that's an assumption.  You didn't use enough punctuation for me to be certain.
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Substitute the values of B and C into the third equation.
A + (3A-9) + (A+B+6) + D = 190
Add
5A - 9 + B + 6 + D = 190
Substitute B again
5A - 9 + (3A-9) + 6 + D = 190
Add/Subtract
8A -12 + D = 190
Add 12 to each side
8A + D = 202
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You seem to be missing some information in the problem.  Would you please check it and resubmit?