SOLUTION: Find three consecutive integers such that the product of the first and second is 36 less than the product of the second and third.
I tried x
x+1
x+2 but did not ge
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-> SOLUTION: Find three consecutive integers such that the product of the first and second is 36 less than the product of the second and third.
I tried x
x+1
x+2 but did not ge
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Question 118447: Find three consecutive integers such that the product of the first and second is 36 less than the product of the second and third.
I tried x
x+1
x+2 but did not get any further :( Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! :
You have your 3 consecutive integers: x, (x+1), (x+2)
:
Find three consecutive integers such that the product of the first and second is 36 less than the product of the second and third.
:
The product of the 1st and 2nd would be:
x(x+1)
:
The product of the 2nd and 3rd would be:
(x+1)(x+2)
:
Write an equation for what it says:
:
Product of 1st & 2nd = Product of 2nd & 3rd - 36
x(x+1) = (x+1)(x+2) - 36
:
FOIL
x^2 + x = x^2 + 3x + 2 - 36
:
Combine all the x's on the left
x^2 - x^2 + x - 3x = 2 - 36
:
-2x = -34
:
x = -34/-2
:
x = +17, our 1st integer
:
I'll let you take it from there. Check your solutions in the equation
