Question 147791: Two positive numbers have a difference of 2 and their reciprocals have a sum of 20/21. What are the numbers? Answer by ptaylor(2198) (Show Source):
You can put this solution on YOUR website! Let x= the larger number; 1/x is the reciprocal
Then x-2=the smaller number; 1/(x-2) is the reciprocal
Now we are told that:
(1/x)+1/(x-2)=20/21 multiply each term by 21x(x-2)
21(x-2)+21x=20x(x-2) get rid of parens
21x-42+21x=20x^2-40x subtract 20x^2 from and add 40x to both sides
21x-42+21x-20x^2+40x=20x^2-20x^2-40x+40x collect like terms
82x-20x^2-42=0 multiply each term by -1 and rearrange
20x^2-82x+42=0 divide each term by 2
10x^2-41x+21=0--------------quadratic in standard form; solve using the quadratic formula: ---------larger number ---------------------smaller number
and ------------------------NO! NEGATIVE #
Now we need to test our answer: (7/2 and 3/2)
2/7+2/3=20/21
6/21 + 14/21=20/21
20/21=20/21
