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

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.