SOLUTION: If the sum of two positive integers is 24 and the difference of their squares is 48. What is the product of the two integers?

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Question 1073860: If the sum of two positive integers is 24 and the difference of their squares is 48. What is the product of the two integers?
Found 2 solutions by ankor@dixie-net.com, ikleyn:
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
If the sum of two positive integers is 24
a + b = 24
a = (-b+24)
and the difference of their squares is 48.
a^2 - b^2 = 48
replace a with (-b+24)
(-b+24)^2 - b^2 = 24
FOIL
b^2 - 48b + 576 - b^2 = 48
-48b = 48 - 576
-48b = -528
b = -528/-48
b = +11
then
a = -11 + 24
b = 13
:
What is the product of the two integers?
11 * 13 = 143
:
;
Check: 13^2 - 11^2 = 48

Answer by ikleyn(52785) About Me  (Show Source):
You can put this solution on YOUR website!
.
The standard method for solving such problems is THIS:

x%5E2+-+y%5E2 = 48,

(x+y)*(x-y) = 48.

Replace x+y by 24 (given !).

24*(x-y) = 48,  ---->

x - y = 2.

Now you have two equations

x + y = 24,
x - y =  2.

Add them and get 2x = 26  --->  x = 13.

Then y = 11.


Answer.  The numbers are 11 and 13.