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Question 228697: Find three consecutive positive integers such that the product of the first and third, minus the second, is 1 more than 7 times the third.
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! Three positive consecutive integers: x, (x+1), (x+2)
:
Find three consecutive positive integers such that the product of the first
and third, minus the second, is 1 more than 7 times the third.
:
x(x+2) - (x+1) = 7(x+2) + 1
:
x^2 + 2x - x - 1 = 7x + 14 + 1
:
x^2 + x - 1 = 7x + 15
:
Combine like terms on the left:
x^2 + x - 7x - 1 - 15 = 0
:
x^2 - 6x - 16 = 0; our old friend, the quadratic equation!
Factors to
(x - 8)(x + 2) = 0
We only want the positive solution in this problem
x = 8, is the 1st integer, then 9, 10; obviously.
;
:
Check our solution in the statement:
"the product of the first and third, minus the second, is 1 more than 7 times the third.
"
8*10 - 9 = 7*10 + 1
80 - 9 = 70 + 1; confirms our solution
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