SOLUTION: Hello, I could really use help with this problem: The product of two numbers is 20, and the sum of their reciprocals is 3/5. Find the numbers. Thank you in advance, Loui

Algebra ->  Sequences-and-series -> SOLUTION: Hello, I could really use help with this problem: The product of two numbers is 20, and the sum of their reciprocals is 3/5. Find the numbers. Thank you in advance, Loui      Log On


   

Question 46085: Hello,
I could really use help with this problem:
The product of two numbers is 20, and the sum of their reciprocals is 3/5. Find the numbers.
Thank you in advance,
Louis
Found 2 solutions by Paul, atif.muhammad:
Answer by Paul(988) About Me  (Show Source):
You can put this solution on YOUR website!
Let the numbers be x and y
EQUATION 1:
xy=20
y=20%2Fx
EQUATION 2:
1%2Fx%2B1%2Fy=3%2F5
SUbsitue for y:
1%2Fx%2B1%2F%2820%2Fx%29=3%2F5
1%2Fx%2Bx%2F20=3%2F5
20%2Bx%5E2=%283%2820%29%28x%29%29%2F5
20%2Bx%5E2=60x%2F5
100%2B5x%5E2=60x
x%5E2-12x%2B20=0
Factor
x=2 and y= 10
Hence, the two numbers are 2 and 10
Paul.

Answer by atif.muhammad(135) About Me  (Show Source):
You can put this solution on YOUR website!
The product of two numbers is 20, and the sum of their reciprocals is 3/5. Find the numbers.

Let one number be x, and the other be y.

Product of two numbers = 20
x * y = 20

xy = 20

Sum of their reciprocals = 3/5

%281%2Fx%29+%2B+%281%2Fy%29+=+%283%2F5%29

Through cross multiplication: %28x%2By%29%2Fxy+=+3%2F5

%28x%2By%29%2Fxy+=+3%2F5

Through substituting xy for 20 (see above, xy=20)

%28x%2By%29%2F20+=+3%2F5

x+y = 12

Now we have a pair of simultaneous equations:

xy = 20 .........(1)
x+y = 12 .......(2)

Manipulate (2)

x +y = 12

y = 12 -x ..........(3)

Substitute (3) into (1):

xy = 20
x(12-x) = 20

12x - x^2 = 20

x^2 -12x +20 = 0

Through factorisation (x-10)(x-2) = 0. Hence, x= 10 or 2

Substitute x=10 or 2 into (3)

y = 12-x

If x = 10, y = 12-10 = 2
If x = 2,  y = 12-2 = 10

So we have two numbers, 10 and 2.