Question

One positive integer is 6 less than twice another. the sum of their squares is 569. find the integers.

Answer 1

Let x be the first integer
Let y be the second integer

x=2y-6

x²+y²=596

(2y-6)²+y²=596
4y²-24y+36+y=569
4y²-23y+36-569=0
4y²-23y-533=0

y= -9.021 or 14.771

I have attached a graph below of the equation.

Since the integer has to be positive (mentioned in question), -9.021 cannot be a solution.

x=2y-6
x=2(14.771)-6
x=23.542

Hope I helped :)