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Find 3 consecutive numbers such that 3 times the first is equal to 8 more than the sum of the other two?
Wiki User
∙ 14y ago
Call the numbers 'x', 'x+1', and 'x+2'.
3x = (x+1) + (x+2) + 8
3x = x + 1 + x + 2 + 8
3x = 2x + 11
Subtract 2x from each side:
x = 11
The numbers are 11, 12, and 13.
===========================
Check:
3(11) ?=? (12 + 13) + 8
33 = 25 + 8
33 = 33
yay !
Wiki User
∙ 14y ago
Anonymous
3,6,9
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