the sum of two numbers is 16 the sum of their reciprocals is 1/3 find the numbers

Question

these question is releated to numbers and please give me full answer

 

Answer 1

Let the 2 numbers be x and y

 

x + y = 16

(1/x) + (1/y) = 1/3

 

x = l16 - y

 

Substitute in the 2nd equation: [1/(16-y)] +(1/y) = 1/3

 

The common denominator on the left side is y(16-y) and the numerator is y + 16 - y = 16

 

16/[y(16-y)]  = 1/3

 

This simplifies to y² - 16y + 48 = 0 which factors as (y-12)(x-4)

 

The roots are symmetric: if you chose y=12, then x=4 and reverse.

 

You should check that the solution is correct by plugging into the equation with the fractions!

 

 

 

 

 

 

 

 

 

 

 

 

the common denominator for the left side is y(16-) and the numerator is y + 16 -y = 16

 

then 16/([y(16-y)] = 1/3 which simplifies to y² - 16y  + 48 = (y-12)(y-4)

 

the roots are symmetric; if ou choose y = 12, then x = 4 and the reverse

 

yo