SOLUTION: The product of second and third of 3 consecutive integers is 3 more than 3 times the second integer. find the 3 integers

Question 534044: The product of second and third of 3 consecutive integers is 3 more than 3 times the second integer. find the 3 integers
Found 2 solutions by boilpoil, MathTherapy:
Answer by boilpoil(127) (Show Source): You can put this solution on YOUR website!
Let x be the first integer
Then x+1 be the second
And x+2 be the third

x=+-2 ---Sorry can't enter the +_ Sign

So the first integer is either 2 or -2

The second integer
=+-2+1
=3/-1

The third integer
=+-3+1
=4/0

Filtering out:
3(4)=3(3)+3
12=9+3
12=12
TRUE

3(-1)=3(0)+3
-3=0+3
-3=3
FALSE

So the three integers are 3, 4 and 5
Answer by MathTherapy(10557) (Show Source): You can put this solution on YOUR website!
The product of second and third of 3 consecutive integers is 3 more than 3 times the second integer. find the 3 integers

Let the 1st integer be F

Then the other 2 are: F + 1, and F + 2

We have: (F + 1)(F + 2) = 3(F + 1) + 3

F, or 1st integer = � 2

Therefore, the consecutive integers are either:, or

------
Check
------
Integers are: 2, 3, & 4
Product of 2nd and 3rd: (3 * 4) = 3(3) + 3 ---- 12 = 9 + 3 ---- 12 = 12 (TRUE)

Integers are: - 2, - 1, & 0
Product of 2nd and 3rd: (- 1 * 0) = 3(- 1) + 3 ---- 0 = - 3 + 3 ---- 0 = 0 (TRUE)

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