Question 182250: find three consective odd intergers such that three time the middle interger isseven more then the sum of the first and thrid interger.
I know x= 1st, x+3 is 2nd and x+4 is thrird.
I keep seting up as
3x=3(x+x+4)+7
Answer by mastermath(14)   (Show Source):
You can put this solution on YOUR website! Let the three consecutive odd integers be:
1st number = x
2nd number = x+2
3rd number = x+4
three time the middle integer is seven more then the sum of the first and third integer.
The sum of the first and the third integer is x+(x+4)= 2x+4.
The three times the middle integer is 3(x+2).
So, the equation becomes
3(x+2) = 7+ (2x+4)
3x + 6 = 7 + 2x + 4
3x + 6 = 11 + 2x
now, subtract 2x on both the sides
3x + 6 - 2x = 11 + 2x - 2x
x + 6 = 11
now, subtract 6 on both the sides
x + 6 - 6 = 11 - 6
x = 5
So, 1st number = x = 5
2nd number = x+2 = 5 + 2 = 7
3rd number = x+4 = 5 + 4 = 9.
Ans: So, the three consecutive odd integers are 5,7 and 9.
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Proof:
Summation of 1st and 3rd number
=5 + 9 = 14
Adding 7 to it we get
14 + 7 = 21.
Three times second number = 7(3) = 21.
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