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