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 :)