SOLUTION: Find two natural numbers, the sum of whose squares is 25 times their sum and also equal to 50 times their difference

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Question 463760: Find two natural numbers, the sum of whose squares is 25 times their sum and also equal to 50 times their difference
Answer by Edwin McCravy(20060)   (Show Source): You can put this solution on YOUR website!
Find two natural numbers, the sum of whose squares is 25 times their sum and also equal to 50 times their difference
Let x be the larger and y be the smaller.

x² + y² = 25(x + y) = 50(x - y)

25(x + y) = 50(x - y) 

Divide both sides by 25

x + y = 2(x - y)

x + y = 2x - 2y

3y = x

Substitute 3y for x in:

x² + y² = 25(x + y)

(3y)² + y² = 25(3y + y)

9y² + y² = 75y + 25y

10y² = 100y

 y² = 10y

y² - 10y = 0

y(y - 10) = 0

y = 0,  y - 10 = 0
             y = 10

0 is not a natural number, so y = 10 

and since

3y = x
 
3(10) = x

30 = x

So the two natural numbers are 30 and 10.

Checking:

The sum of their squares is 30²+10² = 900+100 = 1000

Their sum is 30+10 = 40

25 times 40 = 1000.  That checks.

Their difference is 30-10 = 20

50 times 20 = 1000.  That checks and 
so the answers are indeed 30 and 10.

Edwin

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