SOLUTION: ``excuse me i need help on this question i tried but i got stuck i really need help. The question is the product of the two smallest integers is 5 less then 5 times the larger inte
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Question 827250: ``excuse me i need help on this question i tried but i got stuck i really need help. The question is the product of the two smallest integers is 5 less then 5 times the larger integer.''.
they are 3 consecutive positive intergers
let the smaller interger be represented by x let the larger integer be reprensented by x
thank you Found 2 solutions by stanbon, LinnW:Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! The question is the product of the two smallest integers is 5 less then 5 times the larger integer.''.
they are 3 consecutive positive intergers
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1st: x-1
2nd: x
3rd: x+1
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Equation:
x(x-1) = 5(x+1)-5
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x^2 - x = 5x
----
x^2 -6x = 0
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x(x-6) = 0
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x = 0 or x = 6
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If x = 0 the numbers are -1,0,1
If x = 6 the numbers are 5,6,7
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Cheers,
Stan H.
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You can put this solution on YOUR website! x = first integer
x + 1 = second integer
x + 2 = third integer
The product of the two smallest = five times the largest less 5
(x)(x + 1) = 5(x+2) - 5
x^2 + x = 5x + 10 - 5
x^2 + x = 5x + 5
add -5x -5 to each side.
x^2 -4x -5 = 0
factor
(x - 5)(x + 1) = 0
So x = 5 or x = -1
Assuming we want positive integers,
the three integers are 5, 6 and 7
Let's check by substituting in (x)(x + 1) = 5(x+2) - 5
(5)(5+1) =? 5(5+2) - 5
(5)(6) =? 5(7) - 5
30 =? 35 - 5
30 = 30