SOLUTION: Good day! Kindly help me with this word problem. The denominators of a fraction is 1 more than twice the numerator. If 4 is added to the numerator and subtracted from the deno

Algebra ->  Problems-with-consecutive-odd-even-integers -> SOLUTION: Good day! Kindly help me with this word problem. The denominators of a fraction is 1 more than twice the numerator. If 4 is added to the numerator and subtracted from the deno      Log On


   

Question 683618: Good day! Kindly help me with this word problem.
The denominators of a fraction is 1 more than twice the numerator. If 4 is added to the numerator and subtracted from the denominator, the result is the reciprocal of the fraction. Find the original fraction.
Thank you very much!
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
n%2Fd = the original fraction
;
The denominators of a fraction is 1 more than twice the numerator.
d = 2n + 1
:
If 4 is added to the numerator and subtracted from the denominator, the result is the reciprocal of the fraction.
%28%28n%2B4%29%29%2F%28%28d-4%29%29 = d%2Fn
cross multiply
d(d-4) = n(n+4)
d^2 - 4d = n^2 + 4n
Replace d with (2n+1)
(2n+1)^2 - 4(2n+1) = n^2 + 4n
4n^2 + 4n + 1 - 8n - 4 = n^2 + 4n
4n^2 - 4n - 3 = n^2 + 4n
4n^2 - n^2 - 4n - 4n - 3 = 0
a quadratic equation
3n^2 - 8n - 3 = 0
Factors to
(3n+1)(n-3) = 0
The positive solution is all we want here
n = 3 is the original numerator
and
d = 2(3)+1
d = 7 is the original denominator
:
:
You can check this in the given statement