# The difference of two positive integers is 9 and the difference of their squares is 189. What is the sum of the numbers ?

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let both integerts be n and m

==> n - m = 9 ........(1)

The difference of their square is 189

==> n^2 - m^2 = 189

We knoiw that n^2 - m^2 = (n-m)(n+m)

==> (n-m)(n+m) = 189

but n-m = 0

==> 9(n+m) = 189

Now divide by 9:

==> n+m = 189/9

==: n + m = 21 ...........(2)

Now add (1) and (2):

==> 2n = 30

**==> n= 15 **

==> n+m = 21

==> m= 21- n= 21- 15

**==> m= 6 **

Let the numbers be A and B. Now, the difference of the positive numbers is 9 so A-B =9.

The difference of their squares = A^2 - B^2 = 189.

We also know that A^2 - B^2 = (A - B)(A + B)

=> 189 = (A - B)(A + B)

=> 189 = 9 * (A+B)

=> A + B = 189 / 9

=> A + B = 21

**Therefore the sum of the two numbers is 21.**

Let x and 9-x be the two numbers which has the diffrence of 9

The sum of their square = 189

To find then sum .

x^2+(9-x)^2 = 189.

x^2 +x^2-18x+81 = 189

2x^2 -18x+81-189 = 0

2x^2-18x-108 = 0

x^2 -9x -108 = 0

x = {9+or- sqrt(9^2-4*1(-108))}/2

x1 = (9+3sqrt57)/2

x2 = (9-sqrt57)/2

x1+x2 = 9 obviously.

We'll put x as the larger integer and y the smaller integer.

We'll input the first constraint from enunciation:

x - y = 9 (1)

We'll set the second condition from enunciation:

x^2 - y^2 = 189 (2)

We'll write the difference of squares as a product:

x^2 - y^2 = (x-y)(x+y)

(x-y)(x+y) = 189

But, from (1), x - y = 9

(x-y)(x+y) = 189 => 9(x+y) = 189

We'll divide by 9:

x + y = 21 (3)

We'll add (1) to (3):

x - y + x + y = 9 + 21

We'll eliminate and combine like terms:

2x = 30

We'll divide by 2:

**x = 15**

But, x - y = 9

15 - y = 9

**y = 6**