Question 689065: what are three consecutive numbers such that twice the sum of the first two numbers is 24 more than 3 times the third number. Answer by ReadingBoosters(3246) (Show Source):
You can put this solution on YOUR website! x, x+1, x+2
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2(x+x+1) = 24 + 3(x+2)
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2(2x+1) = 24 + 3x+6
4x + 2 = 24 + 3x + 6
4x - 3x = 24 + 6 - 2
x = 28
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Numbers are , , and
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Check
2(28+29) = 2(57) = 114
3(30) + 24 = 90 + 24 = 114
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