12 is divided into two parts such that the sum of their reciprocals is `12/35` . Find the two parts.

Let the 1^{st} part = x , ∴ 2^{nd} part = 12 - x = `1/x+1/(12-x)=12/35=>((12-x)+x)/(x(12-x))=12/35=>(12-x+x)/(12x-x^2)=12/35` 420 = 144 x -12 x² ⇒ 12 x²-144x +420 x²- 12x +35 = x ²- 7x - 5x + 35 x(x -7) - 5(x -7) x = 7 or 5 ∴ Two parts are 7 and 5

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