SOLUTION: find three consecutive odd integers such that 3 times the sum of all three is 28 more than the product of the first and second integers.
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Question 1029126: find three consecutive odd integers such that 3 times the sum of all three is 28 more than the product of the first and second integers.
Found 2 solutions by Boreal, MathLover1:
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
They are x, x+2 and x+4.
add them to get 3x+6
3(3x+6) -28 = x (x+2)=x^2+2x
9x+18-28=9x-10=x^2+2x
x^2-7x+10=0
(x-5)(x-2)=0
x=5 and 2, but only 5 is odd.
The numbers are 5,7,9
They add to 21
3*21-28=35
That is the product of the first two integers.
Answer by MathLover1(20849) (Show Source): You can put this solution on YOUR website!
Let ,, represent the three consecutive odd integers.
if times the of all three is more than the of the first and second integers, we have:
.....solve for
solutions:
->
or
->->even
so, your numbers are:,,
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