Question 53301
three consecutive integers are such that the first plus one half the second plus seven less than twice the third is 2101.What are the integers?
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Let x = 1st digit; (x+1) = 2nd digit; (x+2) = 3rd digit
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Write an expression for each phrase:
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"first plus one half the second"
     x + .5(x+1)
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"plus seven less that twice the third"
   + 2[x+2] - 7
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Put it all together (the word "is" usually mean equal):
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x + .5(x+1) + 2(x+2)- 7 = 2101
:
x + .5x + .5 + 2x + 4 - 7 = 2101, multiplied what's in brackets
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x + .5x + 2x + .5 + 4 - 7 = 2101, group and combine like terms
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3.5x - 2.5 = 2101
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3.5x = 2101 + 2.5, added 2.5 to both sides
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3.5x = 2103.5
;
x = 2103/3.5, divide both sides by 3.5
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x = 601 is our 1st digit
:
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Check by adding the digits as directed by the problem: 
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601 + .5(602) + 2(603) - 7 =
:
601 + 301 + 1206 - 7 = 2101