Question 63367: There are three consectutive integers such that 8 more than twice the 2nd integer is three times the sum of the 1st and 3rd integer. Find the three integers.
Whomever this may concern, I needed to know if my calculations were correct?
x
x + 2
x + 3
8 + 2 (x+2) = 3 (x+x+3)
8x + 2x + 4 = 6x + 9
2x + 12 = 6x + 9
I then subtract 9 from both sides
2x + 3 = 6x
I then subtract 2x from both sides
+ 3 = 4x
Our teacher provides an answer sheet with our review worksheet. Her answers were 1, 2, and 3.
This is where I am lost; I get 0.75 when I divide 3/4
Thanks for your time.
Answer by 303795(602)   (Show Source):
You can put this solution on YOUR website! Your consecutive integers need to be x, (x + 1) and (x + 2).
The method used seems sound although you ended up with an 8x which should just be an 8 at one stage.
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