I'm assuming that you want an exact answer, not a graphical approximation (which is a lost easier!):
x² + y² = 325 ===> x² + 2xy + y² = 325 + 2xy ===> (x + y)² = 325 + 2·6(x + y)
===> (x + y)² = 325 + 12(x + y) ===> (x + y)² - 12(x + y) - 325 = 0
Factor or use the quadratic formula: [(x + y) - 25][(x + y ) + 13] = 0
===> x = 25 or x = -13
If x + y = 25 ===> y = 25 - x
x² + y² = 325 ===> x² + (25 - x)² = 325 ===> x² + 625 - 50x + x² = 325
===> 2x² - 50x + 300 = 0 ===> x² - 25x + 150 = 0
Factoring: (x - 15)(x - 10) = 0 ===> x = 15 or x = 10
===> when x = 10, y = 15; when x = 15, y = 10
Now, x = -13 gives you two more answers; in a similar solution as the above: x = (-13 ± √481 ) / 2
When x = (-13 + √481 ) / 2, y = (-13 - √481 ) / 2
and when x = (-13 - √481 ) / 2, y = (-13 + √481 ) / 2