Question

Find two numbers such that their sum, their difference, and their product have the ratio of 3:2:5

Answer 1

Answer:

Let the two numbers be x & y:

%28x%2By%29%2F%28x-y%29 = 3%2F2

Cross multiply:

2(x+y) = 3(x-y)

2x + 2y = 3x - 3y

3y + 2y = 3x - 2x

5y = x

:

and

%28x-y%29%2F%28xy%29 = 2%2F5

Cross multiply:

5(x-y) = 2xy

5x - 5y = 2xy

5x - 2xy = 5y

x(5-2y) = 5y

x = 5y%2F%285-2y%29

Substitute 5y for x

5y = 5y%2F%285-2y%29

Just looking at this we can assume

(5-2y) = 1

-2y = 1 - 5

-2y = -4

y = +2

:

Therefore:

x = 5(2)

x = 10

:

Testing the solution: x=10, y=2

12%2F8 = 3%2F2

and

8%2F20 = 2%2F5