SOLUTION: the product of the second and third of three consecutive integers is 2 more than 10 times the first integer?

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Question 381447: the product of the second and third of three consecutive integers is 2 more than 10 times the first integer?
Found 2 solutions by Fombitz, checkley79:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
N, N%2B1, and N%2B2 are the three consecutive integers.
%28N%2B1%29%28N%2B2%29=2%2B10N
N%5E2%2B3N%2BN%2B2=10N%2B2
N%5E2-7N=0
N%28N-7%29=0
Two solutions:
highlight%28N=0%29, then the three integers are 0,1,and 2.
.
.
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N-7=0
highlight_green%28N=7%29, then the three integers are 7, 8, and 9.

Answer by checkley79(3341) About Me  (Show Source):
You can put this solution on YOUR website!
LET X, X+1 & X+2 BE THE 3 INTEGERS.
(X+1)(X+2)=10X+2
X^2+3X+2=10X+2
X^2+3X-10X+2-2=0
X^2-7X=0
X(X-7)=0
X-7=0
X=7 ANS.
PROOF:
(7+1)(7+2)=10*7+2
8*9=70+2
72=72