Question

find two natural no;s ,the sum of whose squares is 25 times their sum and also equal to 50 times their difference

Answer 1

let the numbers be x and y
according to question,
x² + y² = 25(x+y)  ____________(1)
x² + y² = 50(x-y)   ____________(2)

from (1) and (2)
25(x+y) = 50(x-y)
⇒ x+y = 2(x-y) = 2x - 2y
⇒ y+2y = 2x-x
⇒ 3y = x
⇒ x = 3y

putting x in (1)
x² + y² = 25(x+y)
⇒ (3y)² + y² = 25(3y+y)
⇒9y²+y²=25×4y
⇒10y²=100y
⇒y² = 10y
⇒ y = 0 or 10
but y≠0 as y is a natural number

So y = 10
 x = 3y = 30

two numbers are 10 and 30

Answer 2

Let the two numbers be x and y

i.e; x² +y² =25(x+y) (1)

x² +y² =50(x-y) (2)

From 1 & 2

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

x+y =2(x-y)

x+y = 2x - 2y

2y+y = 2x-x

3y = x

Substituting in 1 we get

x²+y² =25( x+y)

(3y)²+y² = 25(3y+y)

9y² +y² =25(4y)

10y² = 100y

y² =10y

y = 10

3y = x = 3*10=30

x= 30 & y=10