SOLUTION: find three consecutive integers such that one half of the first plus one fifth of the second plus one fourth of the third is 33

Question 218818: find three consecutive integers such that one half of the first plus one fifth of the second plus one fourth of the third is 33
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
find three consecutive integers such that one half of the first plus one fifth of the second plus one fourth of the third is 33
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1st: x
2nd: x+1
3rd: x+2
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Equation:
(1/2)x + (1/5)(x+1) + (1/4)(x+2) = 33
Multiply thru by 20 to get:
10x + 4(x+1) + 5(x+2) = 660
19x + 14 = 660
19x = 646
x = 34
Numbers: 34,35,36
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Cheers,
Stan H.
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