SOLUTION: Find 3 consecutive numbers such that twice the smallest is 23 more than the largest.

Question 199262: Find 3 consecutive numbers such that twice the smallest is 23 more than the largest.
Found 2 solutions by stanbon, Earlsdon:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Find 3 consecutive numbers such that twice the smallest is 23 more than the largest.
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1st: x-1
2nd: x
3rd: x+1
---------------------------
Equation:
2(x-1) = x+1 + 23
2x-2 = x + 24
x = 26 (2nd)
x-1= 25 (1st)
x+1=27 (3rd)
==================
Cheers,
Stan H.

Answer by Earlsdon(6294) (Show Source): You can put this solution on YOUR website!
Let x = the first number, (x+1) = the next consecutive number, and (x+2) = the third consecutive number.
According to the problem statement:
Simplify this and solve for x.
Subtract x from both sides.
and..
and...
, so...
The three consecutive numbers are 25, 26, and 27.
Check:
Substitute x = 25.

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