Question

I already solved this problem. I just need help with the HINT part.

Answer 1

$x+y=25 \\ y=25-x~~Other~number~in~terms~of~x \\ \\ x^2+y^2=313 \\ x^2+2xy+y^2=313+2xy \\ (x+y)^2=313+2x(25-x) \\ \frac{25^2-313}{2} =25x-x^2 \\ x^2-25x+156 \\ \\ \Delta =(-25)^2-4\times 1\times 156 \\ \Delta =625-624 \\ \sqrt{\Delta}=1 \\ \\ x= \frac{25 \pm 1}{2} \\ x= \left \{ {{x_1=13} \atop {x_2=12}} \right.$

     The number "x" is given by:
$y= \left \{ {{y_1=25-13 \rightarrow~y_1=12} \atop {y_1=25-12 \rightarrow ~y_1=13}} \right.$
  
     Therefore, we have:

$\boxed {S=({12,13})} \rightarrow~The~curly~braces~does~not~want~appear~here.$