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Radda [10]
3 years ago
14

Find two numbers such that their product, sum and difference are in the ratio 5:4:1'

Mathematics
1 answer:
wel3 years ago
7 0

Answer:

3 1/3 and 2

Step-by-step explanation:

Let the numbers be x and y  and let z be the constant of proportion. Then

xy = 5z  ...............(1)

x + y = 4z ...........(2)

x - y = z ..............(3)     Adding (2) and (3)

2x = 5z

x = 5z/2   ................(4)

Substituting for x in equation 1:

5z/2 *  y = 5z

y = (5z * 2) / 5z = 2.

Substituting for  y in equations (2) and (3):

x + 2 = 4z

x - 2 = z            Subtracting:

4 = 3z

z = 4/3.

Substituting for z in equation (4)

x =  5(4/3) /2

x = 20/6 = 10/3 = 3 1/3.

So the 2 numbers are  3 1/3 and 2.

Checking the results:

xy = 2 * 10/3 = 20/3 = 5 * 4/3 = 5z

x + y = 10/3 + 2 = 16/3 = 4 * 4/3 = 4z.

x - y = 10/3 - 2 = 4/3 = z.

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