SOLUTION: The second of three members are three times the first and the third is four more than one third the first. the sum of this members is 43. find the three numbers

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Question 1065161: The second of three members are three times the first and the third is four more than one third the first. the sum of this members is 43. find the three numbers
Found 2 solutions by JulietG, MathTherapy:
Answer by JulietG(1812)   (Show Source): You can put this solution on YOUR website!
Let the members be represented by A, B, and C.
What do we know from the question?
A + B + C = 43
B = 3A
C = .33A + 4
Substitute the values of the second two equations into the first.
A + B + C = 43
A + (3A) + (.33A+4) = 43
4A + .33A + 4 = 43
4.33A + 4 = 43
4.33A = 39
Divide.
A = 9 (disregard the insignificant places; due to rounding 1/3 to .33)
If we now know that A = 9, we can figure out the others.
B = 3A
B = 3(9)
B = 27
C = .33A + 4
C = 3 + 4
C = 7
Prove it by adding the values together.
9 + 27 + 7 = 43
Success!
Answer by MathTherapy(10555)   (Show Source): You can put this solution on YOUR website!
The second of three members are three times the first and the third is four more than one third the first. the sum of this members is 43. find the three numbers
Let the 1st members number be F

Then 2nd is: 3F, and 3rd is: 	

We then get: 

	

12F + F = 117 ------- Multiplying by LCD, 3	

13F = 117

F, or 1st number = 	

2nd number: 

3rd number:  

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