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

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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) About Me  (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
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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) About Me  (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:
2x+=+%28x%2B2%29%2B23 Simplify this and solve for x.
2x+=+x%2B25 Subtract x from both sides.
highlight_green%28x+=+25%29 and..
highlight_green%28x%2B1+=+26%29 and...
highlight_green%28x%2B2+=+27%29, so...
The three consecutive numbers are 25, 26, and 27.
Check:
2x+=+%28x%2B2%29%2B23 Substitute x = 25.
2%2825%29+=+%2825%2B2%29%2B23
50+=+27%2B23
50+=+50