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Question 197945This question is from textbook beginning and intermediate algebra
: find three consecutive odd integers such that 3 times the sum of all three is 18 more than the product of the first and second integers
This question is from textbook beginning and intermediate algebra
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! find three consecutive odd integers such that 3 times the sum of all three is 18
more than the product of the first and second integers
:
"three consecutive odd integers": x, (x+2), (x+4)
:
"3 times the sum of all three is 18 more than the product of the first and second integers"
3(x + (x+2 + (x+4)) = x(x+2) + 18
:
3(3x + 6) = x^2 + 2x + 18
:
9x + 18 = x^2 + 2x + 18
Subtract 18 from both sides:
9x = x^2 + 2x
:
0 = x^2 + 2x - 9x
:
x^2 - 7x = 0
Factor out x
x(x - 7) = 0
Two solutions
x = 0
and
x = +7
:
Both solutions will work in the original equation, however only x=7, 9, 11 are the odd solutions
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