SOLUTION: I am lost. The sum of the reciprocals of two consecutive even integers is 5/12. Find the two intergers. I dont know where to start, please help.

Algebra ->  Algebra  -> Real-numbers -> SOLUTION: I am lost. The sum of the reciprocals of two consecutive even integers is 5/12. Find the two intergers. I dont know where to start, please help.      Log On

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Question 20856: I am lost. The sum of the reciprocals of two consecutive even integers is 5/12. Find the two intergers. I dont know where to start, please help.
Answer by venugopalramana(3286) About Me  (Show Source):
You can put this solution on YOUR website!
LET ONE EVEN NUMBER BE 2X,ITS NEXT NUMBER WILL BE 2X+2
THEIR RECIPROCALS ARE 1/2X AND 1/2(X+1)
THE SUM OF RECIPROCALS IS 1/2X+1/2(X+1)=5/12
TAKING LCM
(X+1+X)/2(X)(X+1)=5/12
(2X+1)/2X(X+1)=5/12
CROSS MULTIPLYING
12(2X+1)=2*5X(X+1)
24X+12=10X^2+10X
10X^2-14X-12=0
5X^2-7X-6=0
5X^2-10X+3X-6=0
5X(X-2)+3(X-2)=0
(X-2)(5X+3)=0
SO X-2=0
OR X=2
SO THE 2 NUMBERS ARE 2X=4 AND 2X+2=6