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
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.