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. Thank You
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Question 277437: 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. Thank You Answer by checkley77(12844) (Show Source):
You can put this solution on YOUR website! x+y+z=54
x=2y+4 or 2y=x-4 or y=(x-4)/2
z=2x
x+(x-4)/2+2x=54
(2x+x-4+2*2x)/2=54
(2x+x-4+4x)/2=54
(7x-4)/2=54
7x-4=54*2
7x-4=108
7x=108+4
7x=112
x=112/7
x=16 ans.
y=(16-4)/2=12/2=6 ans.
z=2*16=32 ans.
Proof:
16+6+32=54
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