Question 109323
You set it up by writing exactly what it says 
(think of the word "is" as meaning equal "=")
:
Let x = "one positive integer
Let y = "another positive integer
:
One positive integer is 7 less than twice another
   x = 2y - 7
:
the sum of their squares is 346. 
  x^2 + y^2 = 346 
:
Find the integers
:
From the 1st statement, we know that x = (2y-7),
x^2 + y^2 = 346
:
Substitute (2y-7) for x in the above equation
(2y-7)^2 + y^2 = 346
:
FOIL (2y-7)(2y-7)
4y^2 - 28y + 49 + y^2 = 346
:
5y^2 - 28y + 49 - 346 = 0
:
5y^2 - 28y - 297 = 0
:
We probably should use the quadratic formula, however,
we know the positive solution has to be an integer, 
let's play around with the factors of 297, we come up with:
(5y + 27) (y - 11) = 0
:
y = +11
:
x = 2(11) - 7
x = 22 - 7
x = 15
:
Check our solutions:
11^2 + 15^2 =
121 + 225 = 346