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Question 438714: A certain number is 3 times another. If the difference of their reciprocals is 8, find the number. (Choose X as one of the numbers, make an equation and then solve it.)
Answer by sudheerb(5)   (Show Source):
You can put this solution on YOUR website! Let x and y be the two numbers.
Given that certain number is 3 times another.
So, x = 3y ....(1)
Difference of their reciprocals is 8.
(1/x) - (1/y) = 8
Here LCM is xy. Multiply each fraction part with xy.
xy(1/x) - xy(1/y) = 8
(y - x)/xy = 8
Multiply both sides by xy
y - x = 8xy
From (1), we have x = 3y
y - (3y) = 8(3y)y
y - 3y = 8y x 3y
-2y = 24y^2
Add 2y to both sides of equation
-2y + 2y = 24y^2 + 2y
0 = 24y^2 + 2y
24y^2 + 2y = 0
take 2y common from the equation, we get
2y(12y + 1) = 0
Equate each part to zero
2y = 0 or 12y + 1 = 0
y = 0 or 12y = -1
y = 0 or y = -1/12
Therefore y is -1/12, 0.
But here y cannot be 0, because difference of reciprocals of two numbers is 8, then reciprocal of y is infinite which is not satisfying the condition.
so, y = -1/12.
when y = -1/12, then x = 3y = 3(-1/12)= -1/4
Therefore two numbers x and y are -1/4 and -1/12
Regards,
Sudheer(edurite.com)
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