<PRE><B>Question 218818</B>
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. </PRE>