SOLUTION: how do you solve "find three consecutive integers such that twice the smallest is 12 less more than the largest." i know to do Let n=1st CI n+1=2nd CI n+2= 3rd CI But

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Question 469643: how do you solve "find three consecutive integers such that twice the smallest is 12 less more than the largest." i know to do
Let n=1st CI
n+1=2nd CI
n+2= 3rd CI

But how do you get the varbal expression. how do you find twice the smallest and the largest. what is that.
Found 2 solutions by ewatrrr, ccs2011:
Answer by ewatrrr(24785)   (Show Source): You can put this solution on YOUR website!
 

Hi,

Yes! Let n, (n+1) and (n+2) represent the three consecutive integers

Question states***

 2n = (n+2)-12

Solving for x

  n = -10  the three consecutive integers are -10,-9,-8



CHECKING our Answer***

 -20 = -8 - 12 = -20 

Answer by ccs2011(207)   (Show Source): You can put this solution on YOUR website!
 Twice the smallest just means 2 times the lowest integer(n)

12 less means you subtract 12 from the highest integer(n+2)

**Note i think you left something out of the question, Im going to assume its "12 less than 3 times the largest" since that provides the first case in which there is a valid solution**

Equation:

2n = 3(n+2) - 12

2n = 3n - 6

Subtract 3n on both sides

-n = -6

Flip signs

n = 6

Therefore the 3 consecutive integers are 6,7,8 

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