Question 253945: The sum of the reciprocals of two consecutive odd integers is 20/99. Find the two integers. Found 2 solutions by palanisamy, edjones:Answer by palanisamy(496) (Show Source):
You can put this solution on YOUR website! Let the two consecutive odd integers be x and x+2.
Given, (1/x)+(1/x+2) = 20/99
(x+2+x)/x(x+2) = 20/99
99(2x+2) = 20x(x+2)
198x+198 = 20x^2+40x
20x^2+40x-198x-198 = 0
20x^2-158x-198 = 0
10x^2 - 79x-99 = 0
10x^2-90x+11x-99 = 0
10x(x-9)+11(x-9)=0
(x-9)(10x+11) =0
x-9 = 0 or 10x+11 = 0
x = 9 or 10x = -11
x = 9 or x = - 11/10
Since x cannot be a fraction, take x = 9.
So the given numbers are 9 and 11
You can put this solution on YOUR website! Let the numbers be x and x+2
1/x + 1/(x+2)=20/99
99(x+2)+99x=20x(x+2)
99x+198+99x=20x^2+40x
198x+198=20x^2+40x
20x^2-158x-198=0 Subtract the left side from the right.
2(10x^2-79x-99)=0
10*-99=-990. Two factors of -990 whose sum is -79 are -90 and 11
10x^2+11x-90x-99=0
x(10x+11)-9(10x+11)=0 Factoring by grouping.
(x-9)(10x+11)=0
x=9, x=-11/10
.
x=9, x+2=11
.
Ed
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