Asked by rr
There are 3 consecutive even integers such that the quotient obtained by dividing twice the largest integer by the smallest integer is three less than three-fifths of the second integer. What are the integers?
Answers
Answered by
Graham
Three consecutive even integers are:
2n, 2n+2, 2n+4, where n is an integer.
The rule is thus:
2(2n+4)/2n = (3/5)(2n+2) - 3
Rearranging gives:
0 = 6 n<sup>2</sup> - 19n - 20
Find the integer root(s) of this quadratic. Solve for integer n.
Use n to find the three consecutive even numbers.
2n, 2n+2, 2n+4, where n is an integer.
The rule is thus:
2(2n+4)/2n = (3/5)(2n+2) - 3
Rearranging gives:
0 = 6 n<sup>2</sup> - 19n - 20
Find the integer root(s) of this quadratic. Solve for integer n.
Use n to find the three consecutive even numbers.
Answered by
Reiny
I would let my 3 consecutive even integers be
x-2, x, and x+2
2(x+2)/(x-2) =3x/5 - 3
expanding and collecting like terms gives us
3x^2 - 31x + 10 = 0
(x-10)(3x - 1) = 0
x = 10 or x = 1/3, but x is an integer, so
x = 10
the integers are 8, 10 , and 12
x-2, x, and x+2
2(x+2)/(x-2) =3x/5 - 3
expanding and collecting like terms gives us
3x^2 - 31x + 10 = 0
(x-10)(3x - 1) = 0
x = 10 or x = 1/3, but x is an integer, so
x = 10
the integers are 8, 10 , and 12
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