Question

One positive integer is 1 less than twice another. The sum of their squares is 106. Find the integers.

Answer 1

They give us 2 pieces to the puzzle.  Both are positive numbers...x and y.
1.) 1 number is 1 less than twice another number. (x = 2y -1)...and
2.) the sum of their squares is 106.  (x^2 + y^2 = 106).

 substitute the value for x into the second equation.

(2y-1)^2 + y^2 = 106
(2y-1) (2y-1) + y^2 = 106  (use distributive property)
4y^2 - 2y - 2y + 1 + y^2 = 106  (subtract 106 from both sides)
4y^2 - 2y - 2y + 1 + y^2 - 106 = 106 - 106 (combine like terms)
5y^2 - 4y - 105 = 0  (factor)
(y-5) (5y-21) = 0  (set to 0)
y - 5 = 0
y = 5 

substitute the 5 into the equation for y   (x = 2(5) - 1)
           x = 9   if we square 9, we get 81.
subtracted from 106 we have 25...the square root of 25 is 5.

our answers are 5 and 9.