Help me with solving this math question?
2016-12-03 9:26 am
The numerator of a fraction is 2 less than its denominator.When both numerator and denominator are increased by 3,the fraction is increased by 3/20.Find the original fraction.
回答 (2)
2016-12-03 12:13 pm
(x + 1)/(x + 3) - (x - 2)/x = (x - 2)/x + 3/20
(x + 1)/(x + 3) - 2(x - 2)/x = 3/20
20x(x + 1) - 40(x + 3) = 3x(x + 3)
17x^2 - 29x - 120 = 0
(x + 1)/(x + 3) - 2(x - 2)/x = 3/20
20x(x + 1) - 40(x + 3) = 3x(x + 3)
17x^2 - 29x - 120 = 0
2016-12-03 10:08 am
(x-2)/x + 3/20 =(x+1)/ (x+3)
20(x+3)(x-2) +3x(x+3) =20x(x+1)
20(x^2 +x -6) +3x^2 +9x =20x^2 +20x
20x^2 +20x -120 +3x^2 +9x =20x^2 +20x
3x^2 +9x -120 =0
x^2 +3x -40=0
(x+8)(x-5)=0
x=-8 or x=5
original fraction = 3/5
ignore -8
20(x+3)(x-2) +3x(x+3) =20x(x+1)
20(x^2 +x -6) +3x^2 +9x =20x^2 +20x
20x^2 +20x -120 +3x^2 +9x =20x^2 +20x
3x^2 +9x -120 =0
x^2 +3x -40=0
(x+8)(x-5)=0
x=-8 or x=5
original fraction = 3/5
ignore -8
收錄日期: 2021-05-01 20:55:33
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20161203012656AAEGffJ