One positive integer is 7

less than twice another. The sum of their squares is 433. Find the integers

You use a system of equations to solve this.

x will be one integer and y will be the other.

2y - 7 = x

x^2 + y^2 = 433

You can plug (2y - 7) in for x in the second equation.

(2y - 7) ^2 + y^2 = 433

4y^2 - 28y + 49 + y^2 = 433

5y^2 - 28y - 384 = 0

y = 12 or - 32/5

It has to be 12 since - 32/5 is not an integer.

Plug 12 in for y to get x.

2 (12) - 7 = x

x = 17

The integers are 12 and 17.