Question

I want to know the answer

Answer 1

Hello Mate!

Let the positive integers be x and y

$x - y = 5 \\ {x}^{2} + {y}^{2} = 13 \\ {(x - y)}^{2} = {5}^{2} \\ {x}^{2} + {y}^{2} - 2xy = 25 \\ 13 - 2xy = 25 \\ -2xy = 25 - 13 \\ -2xy = 12 \\ xy = \frac{12}{-2} = -6$

Hope it helps☺!✌

Answer 2

★ Hey mate ★

here is your answer !

let the ist digits be x

and second digit be y

condition possible :-

according to question

x - y = 5

or y - x = 5

now ,

sum of their squares =>

x² + y² = 13

( y + 5) ² + y² = 13

=> y² + 25 + 25y + y² = 13

2y² + 25y + 12 = 0

2y² + 24 y + 1y + 12 = 0

2y ( y + 12 ) + 1 ( y + 12 ) = 0

y =- 12 , y = -1 / 2

on taking the value of y = -12

we get

x - y = 5

x +12 = 5

x = -7

or ,

x + 1/2 = 5

2x = 10

x = 5

hence ,

two value of x. = -7 , 5

and , two value of y = -12 , -1 /2



hope it helps you dear !!

$thanks$