Question 828023: Find three consecutive integers such that the sum of the second and the third is half the first decreased by 15.
Answer by alittletooconfused(7)   (Show Source):
You can put this solution on YOUR website! let x = first #; x+1 = second #; x+2 = third #
Consecutive numbers always convert into x, x+1, x+2 and so on depending on how many you need.
sum of the second and the third is half the first decreased by 15
x+1 + x+2 = 1/2x - 15
Combine like terms
x+x + 1+2 = 1/2x - 15
2x + 3 = 1/2x - 15
Subtract 3 from both sides so that only 2x would remain
2x + 3 - 3 = 1/2x - 15 - 3
2x = 1/2x - 18
Multiply both sides by 2 to simply the fraction on the right side.
2(2x) = 2(1/2x - 18)
4x = x - 36
Subtract x from both sides so that only -36 would remain
4x - x = x - 36 - x
3x = -36
Divide both sides by 3 so that only x would remain on the left side
3x/3 = -36/3
x = -12
Substitute:
x = -12
x+1 = -12 + 1 = -11
x+2 = -12 + 2 = -10
Check:
x+1 + x+2 = 1/2x - 15
-11 + -10 = 1/2(-12) - 15
-21 = -6 - 15
-21 = -21
Conclusion: Thus, the three consecutive integers are -12,-11,-10
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