Question

The sum of two numbers is 16 and the sum of their reciprocals is 1/3. Find the numbers.

Answer 1

let one number=x

another number=y

x+y=16

y=16-x

1/x + 1/y=1/3

x+y/xy=1/3

3x+3y=xy

3x+3(16-x)=xy

3x+48-3x=xy

xy=48

x(16-x)=48

-x^2+16x-48=0

-x^2+12x+4x-48=0

-x(x-12)+4(x-12)=0

(-x+4)(x-12)=0

-x+4=0. x-12=0

x=4. x=12

one number=4 or 12

another number= 12 or 4

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