Question

the sum of the two positive integers is 130 and the difference of their squares is 32. the sum of the number is​

Answer 1

Step-by-step explanation:

actually the question you asked is wrong here, it would be the sum of the square of two positive integers

and for that :

x²+y²=130 ------ (i)

x²-y²=32 -----(ii)

2x²=162; x²=81; x=9

put into no (ii) you will get y=7

and the sum of the nos = 7+9=16

Answer 2

Correct Question:

The sum of the squares of two positive integers is 130 and the difference between their squares is 3. Then the sum of the two numbers is?

The correct answer to the question is 16.

i.e., The sum of the two numbers is 16.

Given:

The sum of the squares of two positive integers is 130.

The difference between their squares is 3.

To Find:

The sum of the two numbers is =?

Solution:

Let us assume the two positive integers be 'x' and 'y'.

Now, let us apply the given conditions in mathematical form as follows;

i.e., x² + y² = 130

      x² - y² = 32

Now, On adding the above two equations, we get;

⇒ 2x² = 130 + 32

⇒ 2x² = 162

⇒ x² = 81

⇒ x = ±9

However, since x is a positive integer,

∴ x ≠ -9

i.e., x = 9

Now, Putting the value of x in any of the above equations, we get:

∴ x² + y² = 130

∴ (9²) +  y² = 130

∴ y² = 130 - 81

∴ y² = 49

∴ y = ± 7

But, y is a positive integer,

y ≠ -7; ∴ y = 7

Thus the sum of the two integers is 9 + 7 = 16.

Hence, The sum of the two numbers is 16.

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