SOLUTION: The first of three numbers exceeds twice the second number by 4, while the third number is twice the first. If the sum of the three numbers is 54, find the numbers.

Question 514262: The first of three numbers exceeds twice the second number by 4, while the third number is twice the first. If the sum of the three numbers is 54, find the numbers.
Found 2 solutions by plover, MathTherapy:
Answer by plover(15) (Show Source): You can put this solution on YOUR website!
Let the three numbers be x , y , z
From the data we conclude :
x=2y+4 (1)
z=2x (2)
From (1)
x=2y+4
x-4=2y
or , 2y=x-4
y= (3)
Adding (3) and (2)
z+y=
z+y=
Adding x on both side
x+y+z=
54= given sum of the three numbers is 54
108=7x-4
108+4=7x
x=
=16
From (2)
z=2(16)
=32
y=
=
=6

Answer by MathTherapy(10556) (Show Source): You can put this solution on YOUR website!
The first of three numbers exceeds twice the second number by 4, while the third number is twice the first. If the sum of the three numbers is 54, find the numbers.

Let the second number be S

Then the first number is 2S + 4, and the third is 2(2S + 4), or 4S + 8

Since the three sum to 54, we'll have:

S + 2S + 4 + 4S + 8 = 54

7S + 12 = 54

7S = 42

S, or second number = , or

First number = 2S + 4, or 2(6) + 4, or

Third number = 4S + 8, or 4(6) + 8, or

-------
Check
-------
First (16) exceeds twice second (2 * 6, or 12) by 4 (TRUE)

Third (32) is twice the first, or 16 (TRUE)

Sum of numbers: 6 + 16 + 32 = 54 ------ 54 = 54 (TRUE)

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