Question

One positive integer is 7
less than twice another. The sum of their squares is 433. Find the integers

Answer 1

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.