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

Algebra ->  Problems-with-consecutive-odd-even-integers -> 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      Log On


   

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) About Me  (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) About Me  (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