SOLUTION: 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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Question 53301This question is from textbook introductory algebra
: 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?
This question is from textbook introductory algebra
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
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?
:
Let x = 1st digit; (x+1) = 2nd digit; (x+2) = 3rd digit
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Write an expression for each phrase:
:
"first plus one half the second"
x + .5(x+1)
:
"plus seven less that twice the third"
+ 2[x+2] - 7
:
Put it all together (the word "is" usually mean equal):
:
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:
:
601 + .5(602) + 2(603) - 7 =
:
601 + 301 + 1206 - 7 = 2101
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