SOLUTION: I am thinking of 3 consecutive integers, 4 times the third decreased by one half the second is 4

Question 353692: I am thinking of 3 consecutive integers, 4 times the third decreased by one half the second is 4
Found 2 solutions by stanbon, jsmallt9:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
I am thinking of 3 consecutive integers, 4 times the third decreased by one half the second is 4
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1st: x-1
2nd: x
3rd: x+1
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Equation::
4(x+1) -(1/2) = 4
4x+4 - (1/2) = 4
---
4x = 1/2
x = 1/8
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That equation does not result
in 3 consecutive integers.
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Cheers,
Stan H.
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Answer by jsmallt9(3758) (Show Source): You can put this solution on YOUR website!
Let x = the first integer. Then
x+1 is the second and
x+2 is the third.
"4 times the third decreased by one half the second is 4" now translates into:

Simplifying:

Get rid of the fractions by multiplying both sides by 2:

Subtract 15 fom each side:
7x = -7
Divide both sides by 7:
x = -1
Since "x" was the first integer the others, x+1 and x+2, are 0 and 1
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