the sum of two numbers is 14 & sum of it's reciprocal is 7/20.Find the numbers?

x+y = 14

1/x + 1/y = 7/20

Now just crank it out.

from Steve's ...

1/x + 1/y = 7/20
times 20xy, the LCD
20y + 20x = 7xy
20(y+x) = 7xy
20(14) = 7xy
40 = xy -----> y = 40/x

plug into the first:
x + 40/x = 14
x^2 + 40 - 14x = 0
(x-10)(x-4) = 0
x = 10 or x = 4

if x = 10, y = 40/10 = 4
if x = 4 , y = 40/4 = 10

the numbers are 10 and 4

let x be a number,then 1\x is its reciprocal

A.Q

=x+1\x=7\20
14+1/x=7/20
280x + 20 =7x
280x + 20 - 7x = 0
hence ,answer completed ��

When I answered this question there was another solution after Steve's which has now been deleted.

I only completed Steve's solution to correct that other posting.

To find the numbers, we'll set up a system of equations based on the given information.

Let's assume the two numbers are a and b. According to the problem, we have two pieces of information:

1. The sum of the two numbers is 14, so we can write the equation:
a + b = 14

2. The sum of their reciprocals is 7/20, so the equation becomes:
1/a + 1/b = 7/20

Now, let's solve this system of equations. We can start by multiplying the second equation by the least common multiple (LCM) of a and b to eliminate the denominators. The LCM of a and b is ab, so we have:

b/a + a/b = 7/20

Now, multiply through by ab:

b^2 + a^2 = 7ab/20

Next, rearrange the first equation (a + b = 14) to a = 14 - b and substitute it into the second equation:

b^2 + (14 - b)^2 = 7b(14 - b)/20

Expanding and simplifying:

b^2 + 196 - 28b + b^2 = (98b - 7b^2)/20

Combining like terms:

2b^2 - 28b + 196 = 98b - 7b^2

Moving everything to one side to set the equation to zero:

7b^2 - 126b + 196 = 0

Now, to solve this quadratic equation, we can use the quadratic formula:

b = (-b ± sqrt(b^2 - 4ac)) / (2a)

In this case, a = 7, b = -126, and c = 196. Plugging the values into the formula:

b = (-(-126) ± sqrt((-126)^2 - 4 * 7 * 196)) / (2 * 7)

Simplifying:

b = (126 ± sqrt(12100 - 5488)) / 14

b = (126 ± sqrt(6612)) / 14

Using a calculator, we find that the square root of 6612 is approximately 81.25, so we have:

b = (126 ± 81.25) / 14

Now, we can solve for b:

b1 = (126 + 81.25) / 14 = 207.25 / 14 = 14.8035
b2 = (126 - 81.25) / 14 = 44.75 / 14 = 3.1964

Since we're dealing with real numbers, we'll consider the solutions that are close to natural numbers. In this case, b = 3.1964 is close to 3, so we'll use that.

Substituting b = 3 into the equation a + b = 14, we can solve for a:

a + 3 = 14
a = 14 - 3
a = 11

Therefore, the numbers are 11 and 3.