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Question 123513: Find 3 consecutive multiples of six such that 4 times the first exceeds twice the third by 12.
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! Find 3 consecutive multiples of six such that 4 times the first exceeds twice the third by 12.
:
Let x = 1st multiply of 6
Then
(x+6) = 2nd multiply of 6
and
(x+12) = 3rd multiply of 6
:
Write an equation from the statement,
"4 times the first exceeds twice the third by 12."
:
4x = 2(x+12) + 12
:
4x = 2x + 24 + 12
:
4x - 2x = 36
:
2x = 36
:
x = 18
:
The three multiples of 6 are: 18, 24, 30
:
Check solution in the statement:
"4 times the first exceeds twice the third by 12."
4(18) = 2(30) + 12
72 = 60 + 12
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