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Question 2530: Find three consecutive even integers such that twice the sum of the second and the third exceeds three times the first by 34. All I need is the equation and I can do the rest. I just can't figure it out. Pleasseeeeeeeeeeeeee!
Found 3 solutions by Dymind, longjonsilver, kiru_khandelwal:
Answer by Dymind(3)   (Show Source):
You can put this solution on YOUR website! Let the first even number be x, the numbers are x, x+2, x+4
2(x+2+x+4)=3x+34
2x+4+2x+8=3x+34
4x+12=3x+34
4x-3x=34-12
x=22
Therefore the first number is 22
the second number is 22+2= 24
the third number is 24+2= 26
Answer by longjonsilver(2297)  (Show Source):
Answer by kiru_khandelwal(79) (Show Source):
You can put this solution on YOUR website! Let the three consecutive integers be
x, x+1, x+2
Now, twice the sum of second and the third
=> 2((x+1)+(x+2)) Solving further below
=> 2(2x+3)= 4x + 6
Now, twice the sum of second and the third exceeds three times the first by 34 implies
=> 4x + 6 = 3x + 34
=> 4x - 3x = 34 -6
=> x = 28
So the three consecutive integers are
28,29,30
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